Hypothesis Testing
Hypothesis
The formal testing of hypothesis is a critical step not to be ignored. One of the jobs of statistics is to reduce the uncertainty in our decision making. It does so by helping to identify assumptions and check the validity, in essence, what is really there. It is my personal preference to evaluate stocks or mutual funds to see if they actually exceed the average market return of 10%. The 10% return is the historical return on stocks over the past 75 years, however, each year has its own unique return. When doing so, I am interested in did this mutual fund exceed the 10% annual return. When you look at hypothesis testing there are three ways you can set them up. There are as follows:
HO: = H1: not equal Word In story problem: Is Different
Ho: < or = H1: is greater than Word In story problem: more than, increased
HO: > or = H1: Is less than Word In story problem: decreased, dropped
The hypothesis come in “pre-packaged sets” so you can’t mix and match. So when you are setting them up, look at the alternative (H1) in setting them up. Many times this is the decision that interests you the most. In this case, I wanted returns that exceeded ten percent, so I looked at the alternative hypothesis of greater than when setting up the process. What is helpful in thinking about the hypothesis and alternative hypothesis are the unique qualities of the null or HO, hypotheses. The HO hypothesis has unique characteristics. The unique characteristics is that you assume there to either be “no difference” or status quo. So, I set the hypothesis up as follows:
HO: return is less than or equal to ten percent
H1: return is greater than ten percent
For the hypothesis test, I could have chosen to have it set up two other ways and those are:
HO: the return is equal to ten percent
H1; the return is not equal to ten percent
HO: The return is greater than or equal to ten percent
H1: The return is less than ten percent.
What is helpful in thinking about the hypothesis and alternative hypothesis are the unique qualities of the null or HO, hypotheses. The HO hypothesis has unique characteristics. The unique characteristics are that you assume there to either “no difference” or status quo. In the stating of HO there is always an implied equal. To help you understand this concept lets take a trip to the local vending machine.
Hypothesis Test Thinking and a Vending Machine
Imagine yourself walking up to a vending machine. You want to purchase a bottle of Pepsi for a $1.00. You walk up to the vending machine and assume you are going to put your dollar in the machine, press the Pepsi button, and the machine will dispense the bottle of Pepsi. Do you walk up to the machine, knowing that it is not going to give you your bottle of Pepsi after you insert your dollar? If so, would you proceed with the transaction. No you wouldn’t make the transaction and willingly lose your doll.
Lecture6 Applied Econometrics and Economic Modelingstone55
The manager of a pizza restaurant conducted an experiment to determine if customers prefer a new baking method for pepperoni pizzas. He provided 100 randomly selected customers with both an old-style and new-style pizza and had them rate the difference on a scale from -10 to 10. Based on the customer ratings, the manager wants to use hypothesis testing to determine if he should switch to the new baking method. The null hypothesis is that customers are indifferent between the methods, while the alternative hypothesis is that customers prefer the new method. The results of the experiment provide strong statistical evidence to reject the null hypothesis and support switching to the new baking method.
This document discusses key concepts in applied statistics including hypothesis testing, p-values, types of errors, sensitivity and specificity. It provides examples and explanations of these topics using scenarios about testing if feeding chickens chocolate changes the gender ratio of offspring. Hypothesis testing involves defining the null and alternative hypotheses and using a statistical test to either reject or fail to reject the null hypothesis based on the p-value. Type I and type II errors in hypothesis testing are explained. Sensitivity and specificity in diagnostic tests are introduced using an example about detecting if a car is being stolen.
Chapter 20 and 21 combined testing hypotheses about proportions 2013calculistictt
This document discusses hypotheses testing and the reasoning behind it. It explains that hypotheses testing involves proposing a null hypothesis and an alternative hypothesis based on a parameter of interest. Data is then analyzed to either reject or fail to reject the null hypothesis. Specifically:
1) The null hypothesis proposes a baseline model or value for a parameter.
2) Statistics are calculated based on the data and compared to what we would expect if the null hypothesis is true.
3) If the results are inconsistent enough with the null hypothesis, we can reject it in favor of the alternative hypothesis. Otherwise we fail to reject the null hypothesis.
The goal is to quantify how unlikely the results would be if the null hypothesis is true,
The document provides an introduction to hypothesis testing, including its real-life applications, key definitions, and structure. It defines hypothesis testing as the process of testing the validity of a statistical hypothesis based on a random sample from a population. The document outlines the common steps in hypothesis testing: 1) stating the null and alternative hypotheses, 2) choosing a significance level, 3) determining the test statistic and decision criteria, 4) rejecting or failing to reject the null hypothesis, and 5) drawing a conclusion. It also defines important terminology like population mean, null and alternative hypotheses, test statistic, significance level, critical region, and p-value. Real-life examples from pharmaceutical testing and legal cases are provided to illustrate the motivation for hypothesis
This document summarizes key concepts from a presentation on A/B testing fundamentals. It discusses:
1. The different possible outcomes of A/B tests and how they relate to concepts like true positives, false positives, etc.
2. The difference between false positive rate and false discovery rate. False positive rate considers the probability of a false positive from a single test, while false discovery rate accounts for running multiple tests.
3. How to balance factors like error rates, effect size detection, and test duration by making tradeoffs between them, such as running tests longer to reduce error rates or detect smaller effects.
The document discusses hypothesis testing, which involves testing a hypothesis about a population using a sample of data. It explains that a hypothesis test has four main steps: 1) stating the null and alternative hypotheses, where the null hypothesis asserts there is no difference between the sample and population, 2) setting the significance level, 3) determining the test statistic and critical region for rejecting the null hypothesis, and 4) making a decision to reject or fail to reject the null hypothesis based on whether the test statistic falls in the critical region. Type I and type II errors are also defined. The document provides examples of null and alternative hypotheses using mathematical symbols and discusses how to determine if a p-value is statistically significant.
Chris Stuccio - Data science - Conversion Hotel 2015Webanalisten .nl
Slides of the keynote by Chris Stuccio (USA) at Conversion Hotel 2015, Texel, the Netherlands (#CH2015): "What’s this all about data science? Explain baysian statistics to me as a kid – what should I know?" http://conversionhotel.com
The document discusses confidence interval estimation and hypothesis testing. It introduces key concepts such as point estimates, interval estimates, confidence intervals, null and alternative hypotheses, significance levels, test statistics, decision rules, and p-values. Examples are provided to illustrate how to construct confidence intervals for means and proportions, determine sample sizes, and conduct hypothesis tests for single means using z-tests and t-tests.
Lecture6 Applied Econometrics and Economic Modelingstone55
The manager of a pizza restaurant conducted an experiment to determine if customers prefer a new baking method for pepperoni pizzas. He provided 100 randomly selected customers with both an old-style and new-style pizza and had them rate the difference on a scale from -10 to 10. Based on the customer ratings, the manager wants to use hypothesis testing to determine if he should switch to the new baking method. The null hypothesis is that customers are indifferent between the methods, while the alternative hypothesis is that customers prefer the new method. The results of the experiment provide strong statistical evidence to reject the null hypothesis and support switching to the new baking method.
This document discusses key concepts in applied statistics including hypothesis testing, p-values, types of errors, sensitivity and specificity. It provides examples and explanations of these topics using scenarios about testing if feeding chickens chocolate changes the gender ratio of offspring. Hypothesis testing involves defining the null and alternative hypotheses and using a statistical test to either reject or fail to reject the null hypothesis based on the p-value. Type I and type II errors in hypothesis testing are explained. Sensitivity and specificity in diagnostic tests are introduced using an example about detecting if a car is being stolen.
Chapter 20 and 21 combined testing hypotheses about proportions 2013calculistictt
This document discusses hypotheses testing and the reasoning behind it. It explains that hypotheses testing involves proposing a null hypothesis and an alternative hypothesis based on a parameter of interest. Data is then analyzed to either reject or fail to reject the null hypothesis. Specifically:
1) The null hypothesis proposes a baseline model or value for a parameter.
2) Statistics are calculated based on the data and compared to what we would expect if the null hypothesis is true.
3) If the results are inconsistent enough with the null hypothesis, we can reject it in favor of the alternative hypothesis. Otherwise we fail to reject the null hypothesis.
The goal is to quantify how unlikely the results would be if the null hypothesis is true,
The document provides an introduction to hypothesis testing, including its real-life applications, key definitions, and structure. It defines hypothesis testing as the process of testing the validity of a statistical hypothesis based on a random sample from a population. The document outlines the common steps in hypothesis testing: 1) stating the null and alternative hypotheses, 2) choosing a significance level, 3) determining the test statistic and decision criteria, 4) rejecting or failing to reject the null hypothesis, and 5) drawing a conclusion. It also defines important terminology like population mean, null and alternative hypotheses, test statistic, significance level, critical region, and p-value. Real-life examples from pharmaceutical testing and legal cases are provided to illustrate the motivation for hypothesis
This document summarizes key concepts from a presentation on A/B testing fundamentals. It discusses:
1. The different possible outcomes of A/B tests and how they relate to concepts like true positives, false positives, etc.
2. The difference between false positive rate and false discovery rate. False positive rate considers the probability of a false positive from a single test, while false discovery rate accounts for running multiple tests.
3. How to balance factors like error rates, effect size detection, and test duration by making tradeoffs between them, such as running tests longer to reduce error rates or detect smaller effects.
The document discusses hypothesis testing, which involves testing a hypothesis about a population using a sample of data. It explains that a hypothesis test has four main steps: 1) stating the null and alternative hypotheses, where the null hypothesis asserts there is no difference between the sample and population, 2) setting the significance level, 3) determining the test statistic and critical region for rejecting the null hypothesis, and 4) making a decision to reject or fail to reject the null hypothesis based on whether the test statistic falls in the critical region. Type I and type II errors are also defined. The document provides examples of null and alternative hypotheses using mathematical symbols and discusses how to determine if a p-value is statistically significant.
Chris Stuccio - Data science - Conversion Hotel 2015Webanalisten .nl
Slides of the keynote by Chris Stuccio (USA) at Conversion Hotel 2015, Texel, the Netherlands (#CH2015): "What’s this all about data science? Explain baysian statistics to me as a kid – what should I know?" http://conversionhotel.com
The document discusses confidence interval estimation and hypothesis testing. It introduces key concepts such as point estimates, interval estimates, confidence intervals, null and alternative hypotheses, significance levels, test statistics, decision rules, and p-values. Examples are provided to illustrate how to construct confidence intervals for means and proportions, determine sample sizes, and conduct hypothesis tests for single means using z-tests and t-tests.
Hypothesis testing is a formal statistical procedure used to investigate ideas about the world. There are 5 main steps: 1) state the null and alternate hypotheses, 2) collect representative data, 3) perform an appropriate statistical test comparing within- and between-group variances, 4) decide whether to reject or fail to reject the null based on the p-value, and 5) present the findings. The document provides an example where a t-test found the average height difference between men and women was 13.7cm with a p-value of 0.002, allowing the researcher to reject the null hypothesis that there is no difference.
Hypothesis TestingThe Right HypothesisIn business, or an.docxadampcarr67227
Hypothesis Testing
The Right Hypothesis
In business, or any other discipline, once the question has been asked there must be a statement as to what will or will not occur through testing, measurement, and investigation. This process is known as formulating the right hypothesis. Broadly defined a hypothesis is a statement that the conditions under which something is being measured or evaluated holds true or does not hold true. Further, a business hypothesis is an assumption that is to be tested through market research, data mining, experimental designs, quantitative, and qualitative research endeavors. A hypothesis gives the businessperson a path to follow and specific things to look for along the road.
If the research and statistical data analysis supports and proves the hypothesis that becomes a project well done. If, however, the research data proved a modified version of the hypothesis then re-evaluation for continuation must take place. However, if the research data disproves the hypothesis then the project is usually abandoned.
Hypotheses come in two forms: the null hypothesis and the alternate hypothesis. As a student of applied business statistics you can pick up any number of business statistics textbooks and find a number of opinions on which type of hypothesis should be used in the business world. For the most part, however, and the safest, the better hypothesis to formulate on the basis of the research question asked is what is called the null hypothesis. A null hypothesis states that the research measurement data gathered will not support a difference, relationship, or effect between or amongst those variables being investigated. To the seasoned research investigator having to accept a statement that no differences, relationships, and/or effects will occur based on a statistical data analysis is because when nothing takes place or no differences, effects, or relationship are found there is no possible reason that can be given as to why. This is where most business managers get into trouble when attempting to offer an explanation as to why something has not happened. Attempting to provide an answer to why something has not taken place is akin to discussing how many angels can be placed on the head of a pin—everyone’s answer is plausible and possible. As such business managers need to accept that which has happened and not that which has not happened.
Many business people will skirt the null hypothesis issue by attempting to set analternative hypothesis that states differences, effects and relationships will occur between and amongst that which is being investigated if certain conditions apply.Unfortunately, however, this reverse position is as bad. The research investigator might well be safe if the data analysis detects differences, effect or relationships, but what if it does not? In that case the business manager is back to square one in attempting to explain what has not happened. Although the hypothesis situation may seem c.
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O&M Statistics – Inferential Statistics: Hypothesis Testing
Inferential Statistics
Hypothesis testing
Introduction
In this week, we transition from confidence intervals and interval estimates to hypothesis testing, the basis for inferential statistics. Inferential statistics means using a sample to draw a conclusion about an entire population. A test of hypothesis is a procedure to determine whether sample data provide sufficient evidence to support a position about a population. This position or claim is called the alternative or research hypothesis.
“It is a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement” (Mason & Lind, pg. 336).
This Week in Relation to the Course
Hypothesis testing is at the heart of research. In this week, we examine and practice a procedure to perform tests of hypotheses comparing a sample mean to a population mean and a test of hypotheses comparing two sample means.
The Five-Step Procedure for Hypothesis Testing (you need to show all 5 steps – these contain the same information you would find in a research paper – allows others to see how you arrived at your conclusion and provides a basis for subsequent research).
Step 1
State the null hypothesis – equating the population parameter to a specification. The null hypothesis is always one of status quo or no difference. We call the null hypothesis H0 (H sub zero). It is the hypothesis that contains an equality.
State the alternate hypothesis – The alternate is represented as H1 or HA (H sub one or H sub A). The alternate hypothesis is the exact opposite of the null hypothesis and represents the conclusion supported if the null is rejected. The alternate will not contain an equal sign of the population parameter.
Most of the time, researchers construct tests of hypothesis with the anticipation that the null hypothesis will be rejected.
Step 2
Select a level of significance (α) which will be used when finding critical value(s).
The level you choose (alpha) indicates how confident we wish to be when making the decision.
For example, a .05 alpha level means that we are 95% sure of the reliability of our findings, but there is still a 5% chance of being wrong (what is called the likelihood of committing a Type 1 error).
The level of significance is set by the individual performing the test. Common significance levels are .01, .05, and .10. It is important to always state what the chosen level of significance is.
Step 3
Identify the test statistic – this is the formula you use given the data in the scenario. Simply put, the test statistic may be a Z statistic, a t statistic, or some other distribution. Selection of the correct test statistic will depend on the nature of the data being tested (sample size, whether the population standard deviation is known, whether the data is known to be normally distributed).
The sampling distribution of the test statistic is divided into t.
This document discusses hypothesis testing without statistics using a criminal trial as an example. It explains that in a trial, the jury must decide between a null hypothesis (H0) that the defendant is innocent, and an alternative hypothesis (H1) that the defendant is guilty based on the presented evidence. There are two possible errors - a Type I error of convicting an innocent person, and a Type II error of acquitting a guilty person. The probability of each error is inversely related to the sample size. The document provides examples to illustrate hypothesis testing concepts like rejection regions, test statistics, and interpreting p-values.
The document discusses hypothesis testing and statistical inference. It begins by defining two types of statistical inference - hypothesis testing and parameter estimation. Hypothesis testing determines if sample data is consistent with a hypothesized population parameter, while parameter estimation provides an approximate value of the population parameter.
It then discusses key aspects of hypothesis testing, including stating the null and alternative hypotheses, developing an analysis plan, analyzing sample data, and deciding whether to accept or reject the null hypothesis. Examples are provided to illustrate hypothesis testing methodology and key concepts like p-values, significance levels, directional versus non-directional hypotheses, and applying the steps of hypothesis testing to evaluate a research study's results.
This document discusses hypotheses, hypothesis testing, and research bias. It defines a hypothesis as a tentative assumption about a population parameter that is statistically tested. The main types of hypotheses covered are the null hypothesis (H0), which is attempted to be disproven, and the alternative hypothesis (H1). The four steps of hypothesis testing are outlined as stating the hypotheses, collecting sample data, calculating sample statistics, and analyzing results to accept or reject H0. Key concepts discussed include p-values, levels of significance, type I and type II errors, and bias. Common biases explained are selection, memory/recall, confounding, and interviewer bias.
Okay, let me try to analyze this step-by-step:
1) Null Hypothesis (H0): The advertisement had no effect on sales.
2) Alternative Hypothesis (H1): The advertisement increased sales.
3) We can test this using a paired t-test, since we have sales data from the same shops before and after.
4) Calculate the mean difference between before and after sales for each shop. Then take the average of those differences.
5) Use the t-statistic to determine if the average difference is significantly greater than 0, which would indicate the advertisement increased sales.
So in summary, a paired t-test can be used to determine if the advertisement
Example of a one-sample Z-test The previous lecture in P.docxcravennichole326
Example of a one-sample Z-test
The previous lecture in Powerpoint explained the general procedure for conducting a hypothesis
test. Now let’s go through an example of a hypothesis test to see how it actually works.
Say you’ve developed a new drug that you think might influence people’s cognitive ability,
although you aren’t really sure what it will do. (Yes, this is a really bad study!) You give the
drug to a sample of N = 49 people and then test their IQ. Say their IQ turns out to be X = 106.
Based on decades of test development, we know that in the general U.S. population, the mean IQ
is = 100 with a standard deviation of = 15. So the research question that we want to address
is whether the people who take the “IQ Pill” will have IQ scores that are different from the
general population (suggesting that the pill had some effect), or whether they basically look the
same as everybody else in the country (suggesting that the pill didn’t do anything).
Let’s test this question by using the formal steps of hypothesis testing:
1.Generate H0 and HA
2.Select statistical procedure
3.Select
4.Calculate observed statistic for your data
5.Determine critical statistic
6.Compare (4) and (5)
7.If (4) exceeds (5), reject H0
8.Otherwise, fail to reject H0
1. Generate H0 and HA
So what are H0 and HA? We don’t know whether the drug will make people’s IQs get higher or
lower, so we need to use a 2-tailed or non-directional hypothesis test. Since the mean in the
general population is 100, we will test whether the mean in our test group is equal to 100.
H0: = 100 HA: 100
In this case, refers to the mean of the population that our treatment group represents. This is a
theoretical population of people who have taken our IQ pill. Obviously, this population doesn’t
exist in reality, because the only people who have ever taken our pill are the 49 people who were
in our study. But IN THEORY, a whole lot of other people could also take this drug, and IN
THEORY, they should respond to it in the same way that our 49 participants respond. So we are
making an inference from our 49 participants to that very large group of people who in theory
could also have taken this drug.
2.Select statistical procedure
At this point, I’m just going to tell you that the statistical procedure that you will use is
something called the Z-test. This is a test that is appropriate for testing whether one sample is
different from some specified value when the value of the population standard deviation () is
known. Over the next few days, we will be introduced to some other statistical procedures that
are appropriate for different kinds of situations than this, and you will need to choose which one
is the best to use.
3.Select
In many cases, I will just tell you what level to use. But you should understand why a
researcher might choose one level compared to a.
Example of a one-sample Z-test The previous lecture in P.docxelbanglis
Example of a one-sample Z-test
The previous lecture in Powerpoint explained the general procedure for conducting a hypothesis
test. Now let’s go through an example of a hypothesis test to see how it actually works.
Say you’ve developed a new drug that you think might influence people’s cognitive ability,
although you aren’t really sure what it will do. (Yes, this is a really bad study!) You give the
drug to a sample of N = 49 people and then test their IQ. Say their IQ turns out to be X = 106.
Based on decades of test development, we know that in the general U.S. population, the mean IQ
is = 100 with a standard deviation of = 15. So the research question that we want to address
is whether the people who take the “IQ Pill” will have IQ scores that are different from the
general population (suggesting that the pill had some effect), or whether they basically look the
same as everybody else in the country (suggesting that the pill didn’t do anything).
Let’s test this question by using the formal steps of hypothesis testing:
1.Generate H0 and HA
2.Select statistical procedure
3.Select
4.Calculate observed statistic for your data
5.Determine critical statistic
6.Compare (4) and (5)
7.If (4) exceeds (5), reject H0
8.Otherwise, fail to reject H0
1. Generate H0 and HA
So what are H0 and HA? We don’t know whether the drug will make people’s IQs get higher or
lower, so we need to use a 2-tailed or non-directional hypothesis test. Since the mean in the
general population is 100, we will test whether the mean in our test group is equal to 100.
H0: = 100 HA: 100
In this case, refers to the mean of the population that our treatment group represents. This is a
theoretical population of people who have taken our IQ pill. Obviously, this population doesn’t
exist in reality, because the only people who have ever taken our pill are the 49 people who were
in our study. But IN THEORY, a whole lot of other people could also take this drug, and IN
THEORY, they should respond to it in the same way that our 49 participants respond. So we are
making an inference from our 49 participants to that very large group of people who in theory
could also have taken this drug.
2.Select statistical procedure
At this point, I’m just going to tell you that the statistical procedure that you will use is
something called the Z-test. This is a test that is appropriate for testing whether one sample is
different from some specified value when the value of the population standard deviation () is
known. Over the next few days, we will be introduced to some other statistical procedures that
are appropriate for different kinds of situations than this, and you will need to choose which one
is the best to use.
3.Select
In many cases, I will just tell you what level to use. But you should understand why a
researcher might choose one level compared to a ...
The document discusses key concepts related to sampling methods in marketing research. It defines sampling elements, population, sampling frame, and sampling unit. It presents formulas for calculating sample size when estimating means of continuous variables and proportions. The formula for means involves variables like confidence level (Z), standard deviation (s), and tolerable error (e). The formula for proportions uses variables like confidence level (Z), estimated proportion (p), and tolerable error (e). The document provides an example of each formula and discusses limitations of the formulas related to number of centers, multiple questions, and cell size in analysis.
This presentation will address the issue of sample size determination for social sciences. A simple example is provided for every to understand and explain the sample size determination.
Hypothesis Testing Definitions A statistical hypothesi.docxwilcockiris
Hypothesis Testing
Definitions:
A statistical hypothesis is a guess about a population parameter. The guess may or not be
true.
The null hypothesis, written H0, is a statistical hypothesis that states that there is no
difference between a parameter and a specific value, or that there is no difference between
two parameters.
The alternative hypothesis, written H1 or HA, is a statistical hypothesis that specifies a
specific difference between a parameter and a specific value, or that there is a difference
between two parameters.
Example 1:
A medical researcher is interested in finding out whether a new medication will have
undesirable side effects. She is particularly concerned with the pulse rate of patients who
take the medication. The research question is, will the pulse rate increase, decrease, or
remain the same after a patient takes the medication?
Since the researcher knows that the mean pulse rate for the population under study is 82
beats per minute, the hypotheses for this study are:
H0: µ = 82
HA: µ ≠ 82
The null hypothesis specifies that the mean will remain unchanged and the alternative
hypothesis states that it will be different. This test is called a two-tailed test since the
possible side effects could be to raise or lower the pulse rate. Notice that this is a non
directional hypothesis. The rejection region lies in both tails. We divide the alpha in two
and place half in each tail.
Example 2:
An entrepreneur invents an additive to increase the life of an automobile battery. If the
mean lifetime of the automobile battery is 36 months, then his hypotheses are:
H0: µ ≤ 36
HA: µ > 36
Here, the entrepreneur is only interested in increasing the lifetime of the batteries, so his
alternative hypothesis is that the mean is greater than 36 months. The null hypothesis is
that the mean is less than or equal to 36 months. This test is one-tailed since the interest
is only in an increased lifetime. Notice that the direction of the inequality in the alternate
hypothesis points to the right, same as the area of the curve that forms the rejection
region.
Example 3:
A landlord who wants to lower heating bills in a large apartment complex is considering
using a new type of insulation. If the current average of the monthly heating bills is $78,
his hypotheses about heating costs with the new insulation are:
H0: µ ≥ 78
HA: µ < 78
This test is also a one-tailed test since the landlord is interested only in lowering heating
costs. Notice that the direction of the inequality in the alternate hypothesis points to the
left, same as the area of the curve that forms the rejection region.
Study Design:
After stating the hypotheses, the researcher’s next step is to design the study. In designing
the study, the researcher selects an appropriate statistical test, chooses a level of
significance, and formulates a plan for conducting the study..
Statistics is used to interpret data and draw conclusions about populations based on sample data. Hypothesis testing involves evaluating two statements (the null and alternative hypotheses) about a population using sample data. A hypothesis test determines which statement is best supported.
The key steps in hypothesis testing are to formulate the hypotheses, select an appropriate statistical test, choose a significance level, collect and analyze sample data to calculate a test statistic, determine the probability or critical value associated with the test statistic, and make a decision to reject or fail to reject the null hypothesis based on comparing the probability or test statistic to the significance level and critical value.
An example tests whether the proportion of internet users who shop online is greater than 40% using
Testing of Hypothesis combined with tests.pdfRamBk5
This document discusses hypothesis testing procedures. It defines a hypothesis as a statement about a population parameter that can be tested. The key points covered are:
- Null and alternative hypotheses are defined, with the null hypothesis containing "=", "<", or ">" and the alternative containing "≠", "<", or ">"
- Tests can be one-tailed or two-tailed depending on the alternative hypothesis
- The level of significance and critical values are used to determine whether to reject or fail to reject the null hypothesis
- Type I and type II errors are explained as incorrect rejections or failures to reject the null hypothesis
- Parametric and non-parametric tests are compared based on their data
This document provides an introduction to elementary statistics concepts. It discusses how statistics can be used to answer questions involving experiments with uncertainty, like coin flips or dice rolls. It explains the process of forming a null hypothesis about what would be expected by chance and then using computational simulations to calculate a p-value to determine if the real-world results are statistically significant or likely due just to luck. Key points made are that correlation does not necessarily imply causation, statistical significance does not equal practical importance, and following a proper statistical methodology is better than relying on intuition alone.
This document describes how to perform a chi-square test to determine if two genes are independently assorting or linked. It explains that for a two-point testcross of a heterozygote individual, you expect a 25% ratio for each of the four possible offspring genotypes if the genes are independent. The chi-square test compares observed vs. expected offspring ratios. It notes that the standard test assumes equal segregation of alleles, which may not always be true.
BUS308 – Week 5 Lecture 1 A Different View Expected Ou.docxcurwenmichaela
BUS308 – Week 5 Lecture 1
A Different View
Expected Outcomes
After reading this lecture, the student should be familiar with:
1. What a confidence interval for a statistic is.
2. What a confidence interval for differences is.
3. The difference between statistical and practical significance.
4. The meaning of an Effect Size measure.
Overview
Years ago, a comedy show used to introduce new skits with the phrase “and now for
something completely different.” That seems appropriate for this week’s material.
This week we will look at evaluating our data results in somewhat different ways. One of
the criticisms of the hypothesis testing procedure is that it only shows one value, when it is
reasonably clear that a number of different values would also cause us to reject or not reject a
null hypothesis of no difference. Many managers and researchers would like to see what these
values could be; and, in particular, what are the extreme values as help in making decisions.
Confidence intervals will help us here.
The other criticism of the hypothesis testing procedure is that we can “manage” the
results, or ensure that we will reject the null, by manipulating the sample size. For example, if
we have a difference in a customer preference between two products of only 1%, is this a big
deal? Given the uncertainty contained in sample results, we might tend to think that we can
safely ignore this result. However, if we were to use a sample of, say, 10,000, we would find
that this difference is statistically significant. This, for many, seems to fly in the face of
reasonableness. We will look at a measure of “practical significance,” meaning the likelihood of
the difference being worth paying any attention to, called the effect size to help us here.
Confidence Intervals
A confidence interval is a range of values that, based upon the sample results, most likely
contains the actual population parameter. The “most likely” element is the level of confidence
attached to the interval, 95% confidence interval, 90% confidence interval, 99% confidence
interval, etc. They can be created at any time, with or without performing a statistical test, such
as the t-test.
A confidence interval may be expressed as a range (45 to 51% of the town’s population
support the proposal) or as a mean or proportion with a margin of error (48% of the town
supports the proposal, with a margin of error of 3%). This last format is frequently seen with
opinion poll results, and simply means that you should add and subtract this margin of error from
the reported proportion to obtain the range. With either format, the confidence percent should
also be provided.
Confidence intervals for a single mean (or proportion) are fairly straightforward to
understand, and relate to t-test outcomes simply. Details on how to construct the interval will be
given in this week’s second lecture. We want to understand how to interpret and understa.
This document defines key concepts in hypothesis testing including the null and alternative hypotheses, the five-step hypothesis testing procedure, and types of errors. It provides examples of hypothesis tests for a population mean when the standard deviation is known and unknown, and for a population proportion. The document explains how to set up and conduct hypothesis tests, interpret results, and compute Type I and Type II errors.
Statistical Inference /Hypothesis Testing Dr Nisha Arora
This document discusses hypothesis testing and related statistical concepts. It begins with an outline of the key topics to be covered, including the concept of hypothesis testing, motivation for its use, null and alternative hypotheses, types of errors, and examples. It then provides more detailed explanations and examples of these topics:
- Hypothesis testing involves using sample data to determine if there is evidence to reject or fail to reject claims about a population.
- Examples show how hypothesis testing can help decision makers determine if new suppliers or products meet claimed standards.
- The null hypothesis states there is no difference or effect, while the alternative hypothesis specifies an expected difference.
- Type I errors incorrectly reject the null hypothesis, while type
You are a project manager and believe that your initiative would be .docxadampcarr67227
You are a project manager and believe that your initiative would be more successful if you had a change manager on your team.
Describe
an actual project you have been part of (not necessarily the leader).
Develop
an argument to your manager on the importance of change management.
Describe
the role of a change manager and how it will benefit the project.
Write
a 1,050- word paper using a minimum of two peer-reviewed sources.
Format
your paper consistent with APA guidelines.
.
You are a project manager at a food agricultural organization and yo.docxadampcarr67227
You are a project manager at a food agricultural organization and you are assigned to review nutritional policies.
1). Write the nutritional policies
2). Identify five stakeholders and their roles in the implementation of the nutritional programs at the community level.
.
More Related Content
Similar to Hypothesis TestingHypothesisThe formal testing of hypothesis.docx
Hypothesis testing is a formal statistical procedure used to investigate ideas about the world. There are 5 main steps: 1) state the null and alternate hypotheses, 2) collect representative data, 3) perform an appropriate statistical test comparing within- and between-group variances, 4) decide whether to reject or fail to reject the null based on the p-value, and 5) present the findings. The document provides an example where a t-test found the average height difference between men and women was 13.7cm with a p-value of 0.002, allowing the researcher to reject the null hypothesis that there is no difference.
Hypothesis TestingThe Right HypothesisIn business, or an.docxadampcarr67227
Hypothesis Testing
The Right Hypothesis
In business, or any other discipline, once the question has been asked there must be a statement as to what will or will not occur through testing, measurement, and investigation. This process is known as formulating the right hypothesis. Broadly defined a hypothesis is a statement that the conditions under which something is being measured or evaluated holds true or does not hold true. Further, a business hypothesis is an assumption that is to be tested through market research, data mining, experimental designs, quantitative, and qualitative research endeavors. A hypothesis gives the businessperson a path to follow and specific things to look for along the road.
If the research and statistical data analysis supports and proves the hypothesis that becomes a project well done. If, however, the research data proved a modified version of the hypothesis then re-evaluation for continuation must take place. However, if the research data disproves the hypothesis then the project is usually abandoned.
Hypotheses come in two forms: the null hypothesis and the alternate hypothesis. As a student of applied business statistics you can pick up any number of business statistics textbooks and find a number of opinions on which type of hypothesis should be used in the business world. For the most part, however, and the safest, the better hypothesis to formulate on the basis of the research question asked is what is called the null hypothesis. A null hypothesis states that the research measurement data gathered will not support a difference, relationship, or effect between or amongst those variables being investigated. To the seasoned research investigator having to accept a statement that no differences, relationships, and/or effects will occur based on a statistical data analysis is because when nothing takes place or no differences, effects, or relationship are found there is no possible reason that can be given as to why. This is where most business managers get into trouble when attempting to offer an explanation as to why something has not happened. Attempting to provide an answer to why something has not taken place is akin to discussing how many angels can be placed on the head of a pin—everyone’s answer is plausible and possible. As such business managers need to accept that which has happened and not that which has not happened.
Many business people will skirt the null hypothesis issue by attempting to set analternative hypothesis that states differences, effects and relationships will occur between and amongst that which is being investigated if certain conditions apply.Unfortunately, however, this reverse position is as bad. The research investigator might well be safe if the data analysis detects differences, effect or relationships, but what if it does not? In that case the business manager is back to square one in attempting to explain what has not happened. Although the hypothesis situation may seem c.
PAGE
O&M Statistics – Inferential Statistics: Hypothesis Testing
Inferential Statistics
Hypothesis testing
Introduction
In this week, we transition from confidence intervals and interval estimates to hypothesis testing, the basis for inferential statistics. Inferential statistics means using a sample to draw a conclusion about an entire population. A test of hypothesis is a procedure to determine whether sample data provide sufficient evidence to support a position about a population. This position or claim is called the alternative or research hypothesis.
“It is a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement” (Mason & Lind, pg. 336).
This Week in Relation to the Course
Hypothesis testing is at the heart of research. In this week, we examine and practice a procedure to perform tests of hypotheses comparing a sample mean to a population mean and a test of hypotheses comparing two sample means.
The Five-Step Procedure for Hypothesis Testing (you need to show all 5 steps – these contain the same information you would find in a research paper – allows others to see how you arrived at your conclusion and provides a basis for subsequent research).
Step 1
State the null hypothesis – equating the population parameter to a specification. The null hypothesis is always one of status quo or no difference. We call the null hypothesis H0 (H sub zero). It is the hypothesis that contains an equality.
State the alternate hypothesis – The alternate is represented as H1 or HA (H sub one or H sub A). The alternate hypothesis is the exact opposite of the null hypothesis and represents the conclusion supported if the null is rejected. The alternate will not contain an equal sign of the population parameter.
Most of the time, researchers construct tests of hypothesis with the anticipation that the null hypothesis will be rejected.
Step 2
Select a level of significance (α) which will be used when finding critical value(s).
The level you choose (alpha) indicates how confident we wish to be when making the decision.
For example, a .05 alpha level means that we are 95% sure of the reliability of our findings, but there is still a 5% chance of being wrong (what is called the likelihood of committing a Type 1 error).
The level of significance is set by the individual performing the test. Common significance levels are .01, .05, and .10. It is important to always state what the chosen level of significance is.
Step 3
Identify the test statistic – this is the formula you use given the data in the scenario. Simply put, the test statistic may be a Z statistic, a t statistic, or some other distribution. Selection of the correct test statistic will depend on the nature of the data being tested (sample size, whether the population standard deviation is known, whether the data is known to be normally distributed).
The sampling distribution of the test statistic is divided into t.
This document discusses hypothesis testing without statistics using a criminal trial as an example. It explains that in a trial, the jury must decide between a null hypothesis (H0) that the defendant is innocent, and an alternative hypothesis (H1) that the defendant is guilty based on the presented evidence. There are two possible errors - a Type I error of convicting an innocent person, and a Type II error of acquitting a guilty person. The probability of each error is inversely related to the sample size. The document provides examples to illustrate hypothesis testing concepts like rejection regions, test statistics, and interpreting p-values.
The document discusses hypothesis testing and statistical inference. It begins by defining two types of statistical inference - hypothesis testing and parameter estimation. Hypothesis testing determines if sample data is consistent with a hypothesized population parameter, while parameter estimation provides an approximate value of the population parameter.
It then discusses key aspects of hypothesis testing, including stating the null and alternative hypotheses, developing an analysis plan, analyzing sample data, and deciding whether to accept or reject the null hypothesis. Examples are provided to illustrate hypothesis testing methodology and key concepts like p-values, significance levels, directional versus non-directional hypotheses, and applying the steps of hypothesis testing to evaluate a research study's results.
This document discusses hypotheses, hypothesis testing, and research bias. It defines a hypothesis as a tentative assumption about a population parameter that is statistically tested. The main types of hypotheses covered are the null hypothesis (H0), which is attempted to be disproven, and the alternative hypothesis (H1). The four steps of hypothesis testing are outlined as stating the hypotheses, collecting sample data, calculating sample statistics, and analyzing results to accept or reject H0. Key concepts discussed include p-values, levels of significance, type I and type II errors, and bias. Common biases explained are selection, memory/recall, confounding, and interviewer bias.
Okay, let me try to analyze this step-by-step:
1) Null Hypothesis (H0): The advertisement had no effect on sales.
2) Alternative Hypothesis (H1): The advertisement increased sales.
3) We can test this using a paired t-test, since we have sales data from the same shops before and after.
4) Calculate the mean difference between before and after sales for each shop. Then take the average of those differences.
5) Use the t-statistic to determine if the average difference is significantly greater than 0, which would indicate the advertisement increased sales.
So in summary, a paired t-test can be used to determine if the advertisement
Example of a one-sample Z-test The previous lecture in P.docxcravennichole326
Example of a one-sample Z-test
The previous lecture in Powerpoint explained the general procedure for conducting a hypothesis
test. Now let’s go through an example of a hypothesis test to see how it actually works.
Say you’ve developed a new drug that you think might influence people’s cognitive ability,
although you aren’t really sure what it will do. (Yes, this is a really bad study!) You give the
drug to a sample of N = 49 people and then test their IQ. Say their IQ turns out to be X = 106.
Based on decades of test development, we know that in the general U.S. population, the mean IQ
is = 100 with a standard deviation of = 15. So the research question that we want to address
is whether the people who take the “IQ Pill” will have IQ scores that are different from the
general population (suggesting that the pill had some effect), or whether they basically look the
same as everybody else in the country (suggesting that the pill didn’t do anything).
Let’s test this question by using the formal steps of hypothesis testing:
1.Generate H0 and HA
2.Select statistical procedure
3.Select
4.Calculate observed statistic for your data
5.Determine critical statistic
6.Compare (4) and (5)
7.If (4) exceeds (5), reject H0
8.Otherwise, fail to reject H0
1. Generate H0 and HA
So what are H0 and HA? We don’t know whether the drug will make people’s IQs get higher or
lower, so we need to use a 2-tailed or non-directional hypothesis test. Since the mean in the
general population is 100, we will test whether the mean in our test group is equal to 100.
H0: = 100 HA: 100
In this case, refers to the mean of the population that our treatment group represents. This is a
theoretical population of people who have taken our IQ pill. Obviously, this population doesn’t
exist in reality, because the only people who have ever taken our pill are the 49 people who were
in our study. But IN THEORY, a whole lot of other people could also take this drug, and IN
THEORY, they should respond to it in the same way that our 49 participants respond. So we are
making an inference from our 49 participants to that very large group of people who in theory
could also have taken this drug.
2.Select statistical procedure
At this point, I’m just going to tell you that the statistical procedure that you will use is
something called the Z-test. This is a test that is appropriate for testing whether one sample is
different from some specified value when the value of the population standard deviation () is
known. Over the next few days, we will be introduced to some other statistical procedures that
are appropriate for different kinds of situations than this, and you will need to choose which one
is the best to use.
3.Select
In many cases, I will just tell you what level to use. But you should understand why a
researcher might choose one level compared to a.
Example of a one-sample Z-test The previous lecture in P.docxelbanglis
Example of a one-sample Z-test
The previous lecture in Powerpoint explained the general procedure for conducting a hypothesis
test. Now let’s go through an example of a hypothesis test to see how it actually works.
Say you’ve developed a new drug that you think might influence people’s cognitive ability,
although you aren’t really sure what it will do. (Yes, this is a really bad study!) You give the
drug to a sample of N = 49 people and then test their IQ. Say their IQ turns out to be X = 106.
Based on decades of test development, we know that in the general U.S. population, the mean IQ
is = 100 with a standard deviation of = 15. So the research question that we want to address
is whether the people who take the “IQ Pill” will have IQ scores that are different from the
general population (suggesting that the pill had some effect), or whether they basically look the
same as everybody else in the country (suggesting that the pill didn’t do anything).
Let’s test this question by using the formal steps of hypothesis testing:
1.Generate H0 and HA
2.Select statistical procedure
3.Select
4.Calculate observed statistic for your data
5.Determine critical statistic
6.Compare (4) and (5)
7.If (4) exceeds (5), reject H0
8.Otherwise, fail to reject H0
1. Generate H0 and HA
So what are H0 and HA? We don’t know whether the drug will make people’s IQs get higher or
lower, so we need to use a 2-tailed or non-directional hypothesis test. Since the mean in the
general population is 100, we will test whether the mean in our test group is equal to 100.
H0: = 100 HA: 100
In this case, refers to the mean of the population that our treatment group represents. This is a
theoretical population of people who have taken our IQ pill. Obviously, this population doesn’t
exist in reality, because the only people who have ever taken our pill are the 49 people who were
in our study. But IN THEORY, a whole lot of other people could also take this drug, and IN
THEORY, they should respond to it in the same way that our 49 participants respond. So we are
making an inference from our 49 participants to that very large group of people who in theory
could also have taken this drug.
2.Select statistical procedure
At this point, I’m just going to tell you that the statistical procedure that you will use is
something called the Z-test. This is a test that is appropriate for testing whether one sample is
different from some specified value when the value of the population standard deviation () is
known. Over the next few days, we will be introduced to some other statistical procedures that
are appropriate for different kinds of situations than this, and you will need to choose which one
is the best to use.
3.Select
In many cases, I will just tell you what level to use. But you should understand why a
researcher might choose one level compared to a ...
The document discusses key concepts related to sampling methods in marketing research. It defines sampling elements, population, sampling frame, and sampling unit. It presents formulas for calculating sample size when estimating means of continuous variables and proportions. The formula for means involves variables like confidence level (Z), standard deviation (s), and tolerable error (e). The formula for proportions uses variables like confidence level (Z), estimated proportion (p), and tolerable error (e). The document provides an example of each formula and discusses limitations of the formulas related to number of centers, multiple questions, and cell size in analysis.
This presentation will address the issue of sample size determination for social sciences. A simple example is provided for every to understand and explain the sample size determination.
Hypothesis Testing Definitions A statistical hypothesi.docxwilcockiris
Hypothesis Testing
Definitions:
A statistical hypothesis is a guess about a population parameter. The guess may or not be
true.
The null hypothesis, written H0, is a statistical hypothesis that states that there is no
difference between a parameter and a specific value, or that there is no difference between
two parameters.
The alternative hypothesis, written H1 or HA, is a statistical hypothesis that specifies a
specific difference between a parameter and a specific value, or that there is a difference
between two parameters.
Example 1:
A medical researcher is interested in finding out whether a new medication will have
undesirable side effects. She is particularly concerned with the pulse rate of patients who
take the medication. The research question is, will the pulse rate increase, decrease, or
remain the same after a patient takes the medication?
Since the researcher knows that the mean pulse rate for the population under study is 82
beats per minute, the hypotheses for this study are:
H0: µ = 82
HA: µ ≠ 82
The null hypothesis specifies that the mean will remain unchanged and the alternative
hypothesis states that it will be different. This test is called a two-tailed test since the
possible side effects could be to raise or lower the pulse rate. Notice that this is a non
directional hypothesis. The rejection region lies in both tails. We divide the alpha in two
and place half in each tail.
Example 2:
An entrepreneur invents an additive to increase the life of an automobile battery. If the
mean lifetime of the automobile battery is 36 months, then his hypotheses are:
H0: µ ≤ 36
HA: µ > 36
Here, the entrepreneur is only interested in increasing the lifetime of the batteries, so his
alternative hypothesis is that the mean is greater than 36 months. The null hypothesis is
that the mean is less than or equal to 36 months. This test is one-tailed since the interest
is only in an increased lifetime. Notice that the direction of the inequality in the alternate
hypothesis points to the right, same as the area of the curve that forms the rejection
region.
Example 3:
A landlord who wants to lower heating bills in a large apartment complex is considering
using a new type of insulation. If the current average of the monthly heating bills is $78,
his hypotheses about heating costs with the new insulation are:
H0: µ ≥ 78
HA: µ < 78
This test is also a one-tailed test since the landlord is interested only in lowering heating
costs. Notice that the direction of the inequality in the alternate hypothesis points to the
left, same as the area of the curve that forms the rejection region.
Study Design:
After stating the hypotheses, the researcher’s next step is to design the study. In designing
the study, the researcher selects an appropriate statistical test, chooses a level of
significance, and formulates a plan for conducting the study..
Statistics is used to interpret data and draw conclusions about populations based on sample data. Hypothesis testing involves evaluating two statements (the null and alternative hypotheses) about a population using sample data. A hypothesis test determines which statement is best supported.
The key steps in hypothesis testing are to formulate the hypotheses, select an appropriate statistical test, choose a significance level, collect and analyze sample data to calculate a test statistic, determine the probability or critical value associated with the test statistic, and make a decision to reject or fail to reject the null hypothesis based on comparing the probability or test statistic to the significance level and critical value.
An example tests whether the proportion of internet users who shop online is greater than 40% using
Testing of Hypothesis combined with tests.pdfRamBk5
This document discusses hypothesis testing procedures. It defines a hypothesis as a statement about a population parameter that can be tested. The key points covered are:
- Null and alternative hypotheses are defined, with the null hypothesis containing "=", "<", or ">" and the alternative containing "≠", "<", or ">"
- Tests can be one-tailed or two-tailed depending on the alternative hypothesis
- The level of significance and critical values are used to determine whether to reject or fail to reject the null hypothesis
- Type I and type II errors are explained as incorrect rejections or failures to reject the null hypothesis
- Parametric and non-parametric tests are compared based on their data
This document provides an introduction to elementary statistics concepts. It discusses how statistics can be used to answer questions involving experiments with uncertainty, like coin flips or dice rolls. It explains the process of forming a null hypothesis about what would be expected by chance and then using computational simulations to calculate a p-value to determine if the real-world results are statistically significant or likely due just to luck. Key points made are that correlation does not necessarily imply causation, statistical significance does not equal practical importance, and following a proper statistical methodology is better than relying on intuition alone.
This document describes how to perform a chi-square test to determine if two genes are independently assorting or linked. It explains that for a two-point testcross of a heterozygote individual, you expect a 25% ratio for each of the four possible offspring genotypes if the genes are independent. The chi-square test compares observed vs. expected offspring ratios. It notes that the standard test assumes equal segregation of alleles, which may not always be true.
BUS308 – Week 5 Lecture 1 A Different View Expected Ou.docxcurwenmichaela
BUS308 – Week 5 Lecture 1
A Different View
Expected Outcomes
After reading this lecture, the student should be familiar with:
1. What a confidence interval for a statistic is.
2. What a confidence interval for differences is.
3. The difference between statistical and practical significance.
4. The meaning of an Effect Size measure.
Overview
Years ago, a comedy show used to introduce new skits with the phrase “and now for
something completely different.” That seems appropriate for this week’s material.
This week we will look at evaluating our data results in somewhat different ways. One of
the criticisms of the hypothesis testing procedure is that it only shows one value, when it is
reasonably clear that a number of different values would also cause us to reject or not reject a
null hypothesis of no difference. Many managers and researchers would like to see what these
values could be; and, in particular, what are the extreme values as help in making decisions.
Confidence intervals will help us here.
The other criticism of the hypothesis testing procedure is that we can “manage” the
results, or ensure that we will reject the null, by manipulating the sample size. For example, if
we have a difference in a customer preference between two products of only 1%, is this a big
deal? Given the uncertainty contained in sample results, we might tend to think that we can
safely ignore this result. However, if we were to use a sample of, say, 10,000, we would find
that this difference is statistically significant. This, for many, seems to fly in the face of
reasonableness. We will look at a measure of “practical significance,” meaning the likelihood of
the difference being worth paying any attention to, called the effect size to help us here.
Confidence Intervals
A confidence interval is a range of values that, based upon the sample results, most likely
contains the actual population parameter. The “most likely” element is the level of confidence
attached to the interval, 95% confidence interval, 90% confidence interval, 99% confidence
interval, etc. They can be created at any time, with or without performing a statistical test, such
as the t-test.
A confidence interval may be expressed as a range (45 to 51% of the town’s population
support the proposal) or as a mean or proportion with a margin of error (48% of the town
supports the proposal, with a margin of error of 3%). This last format is frequently seen with
opinion poll results, and simply means that you should add and subtract this margin of error from
the reported proportion to obtain the range. With either format, the confidence percent should
also be provided.
Confidence intervals for a single mean (or proportion) are fairly straightforward to
understand, and relate to t-test outcomes simply. Details on how to construct the interval will be
given in this week’s second lecture. We want to understand how to interpret and understa.
This document defines key concepts in hypothesis testing including the null and alternative hypotheses, the five-step hypothesis testing procedure, and types of errors. It provides examples of hypothesis tests for a population mean when the standard deviation is known and unknown, and for a population proportion. The document explains how to set up and conduct hypothesis tests, interpret results, and compute Type I and Type II errors.
Statistical Inference /Hypothesis Testing Dr Nisha Arora
This document discusses hypothesis testing and related statistical concepts. It begins with an outline of the key topics to be covered, including the concept of hypothesis testing, motivation for its use, null and alternative hypotheses, types of errors, and examples. It then provides more detailed explanations and examples of these topics:
- Hypothesis testing involves using sample data to determine if there is evidence to reject or fail to reject claims about a population.
- Examples show how hypothesis testing can help decision makers determine if new suppliers or products meet claimed standards.
- The null hypothesis states there is no difference or effect, while the alternative hypothesis specifies an expected difference.
- Type I errors incorrectly reject the null hypothesis, while type
You are a project manager and believe that your initiative would be .docxadampcarr67227
You are a project manager and believe that your initiative would be more successful if you had a change manager on your team.
Describe
an actual project you have been part of (not necessarily the leader).
Develop
an argument to your manager on the importance of change management.
Describe
the role of a change manager and how it will benefit the project.
Write
a 1,050- word paper using a minimum of two peer-reviewed sources.
Format
your paper consistent with APA guidelines.
.
You are a project manager at a food agricultural organization and yo.docxadampcarr67227
You are a project manager at a food agricultural organization and you are assigned to review nutritional policies.
1). Write the nutritional policies
2). Identify five stakeholders and their roles in the implementation of the nutritional programs at the community level.
.
You are a nursing educator and you are given an assignment to teach .docxadampcarr67227
You are a nursing educator and you are given an assignment to teach a RN/LPN NCLEX review course.
Please develop a complete review course power point presentation with detail speaker notes that will be used to teach the review in its entirely. You want student to pass the nclex exam on the first try. please rearrange order and at to it as you deem fit if I left out some thing (please insert pictures and diagram to enhance lecture) Please be very creative and colorful (Presentation to be shown to a large audience. Please be very detail but highlighting the most important detail.
The power points must include elements as follow:
1. nclex question types
2. steps of question analysis
3. critical thinking and rewording
4. how to dissect nclex question
5. what are considered hig level questions
6. deciding what is important
7. looking for patterns and relationships
8. identifying the problem
9. transferring knowledge from one situation to another
10. applying knowledge
11. discriminating between possible choices and/or course of action
12. evaluating according to criteria established
13. eliminating incorrect answer choices
14. strategies for alternate formate question: select all that apply
15. solving alternate formate questions: select all that apply.
16. prioritization
17. delegation
18. safety and infection control
19. maslow's hierarchy of needs
20. how to approach psychosocial condition question
21. how to answer psych questions
22. how to identify psych diagnosis and nursing care of the psychiatric patient
how to answer health promotion and maintenance question
23. tips on how to pass nclex exam
24. hot spot questions and how to solve them
25. fill in blank question and how to solve them and select all that apply
drag and drop question and how to solve them
26. tips on how to analyze a question
27. NURSING LAB VALUES TO KNOW
28. NURSING DRUGS TO KNOW AND LEVELS
INFORMATION ON THE FOLLOWING(with nursing most important intervention and things to watch for/ complication problems up each system)
Care of the pediatric patient
Care of OB (maternity) patient
Care of a pre-op patient
Care of a patient post op
Care of a respiratory patient
Care of a cardiac patient
Care of a gastro/intestinal patient
Care of caner patient
Care of urinary system patient
endoceine system
liver
pancreas
nutritional problem
chronic neurological problems
stroke
intracranial problems
muscle skeletal problems
emergency, terrorism and disaster nursing
fluid and electrolytes
the different in IV solution
Administering Blood
Conscious sedation
Reproductive system
nutrition for a newborn
drug calculation
Immunization when due and side effect
Kidney disorders and care of a renal patient with labs
Diabetes management
spinal cord injury
musculoskeletal problem
alzheimer's disease
ABG interpetation
drug calculation
oxygen supplement and delivery system
integumentary system
bur.
You are a paralegal working at law office of James Adams, Esq. On No.docxadampcarr67227
You are a paralegal working at law office of James Adams, Esq. On November 10, 2010, Adams is assigned by the court to represent John Edwinson, against whom a paternity petition has been filed. There is a hearing scheduled for march 13, 2011. Edwinson is not a cooperative client. He frequent misses appointment at the law firm office. Frustrated, Adams sends Edwinson a short letter on March 1,2011 that says, " Due to your noncooperation, I am withdrawing from the case as your representative effective immediately." Any ethical problem
.
you are a paralegal working at the law office of Smith & Smith. The .docxadampcarr67227
you are a paralegal working at the law office of Smith & Smith. The office represents David Gerry in a divorce action against his wife, Lena Gerry. One of the disputes is how to divide business assets acquired during the marriage. In an effort to pressure Lena to divide the assets in his favor, David tells his attorney to request sole physical and legal custody of their children even though David has no desire to raise the children. He knows, however, that Lena is terrified at the thought of losing sole custody herself. David wants his attorney to engage in extensive discovery (depositions, interrogatories, etc.) On the custody issue for the sole purpose of wearing Lena down in hope that she will reduce her claims on the business assets. Any ethical problems?
.
You are a police officer who has been selected to participate in a p.docxadampcarr67227
You are a police officer who has been selected to participate in a public relations task force to address a growing problem: the negative public perception of the police.
The media has been tough on departments around the city, and the police chief wants to address the issue head on. You just completed the first task force meeting, and the facilitator wants you to present information and recommendations regarding how to change the public’s perception.
Create
an 8- to 10-slide Microsoft® PowerPoint® presentation in which you:
Explain how an inductive fallacy (e.g., generalizations, weak analogy) or a fallacy of language (e.g., confusing explanations) may affect the public perception of the police.
Provide a categorical claim related to the negative public perception of the police.
Create a visual showing a categorical relation that is negative between the police and the public.
Provide recommendations and examples about what the department can do to:
Change the perception
Develop a positive relationship with the public.
Include
comprehensive speaker notes.
Cite
at least 1 reference to support your assignment.
Format
your citations according to APA guidelines
.
You are a newly-minted, tax-paying and law-abiding, permanent res.docxadampcarr67227
"You are a newly-minted, tax-paying and law-abiding, permanent resident of Canada.
In the context of the Canadian multicultural society, you are involved in your community, holding a volunteer office (e.g. VP, Secretary etc.) in your community association.
At the last community meeting several members raised the issue of whether what is going on the Canadian political scene, such as:
the Jody Wilson- Raybould, former federal Justice Minister and Attorney General, story
the Bill Morneau, former federal Minister of Finance, story, and especially
the Julie Payette, former Governor General of Canada, story
are indicative of changes, in the Canadian society, which will impact the country and its communities.
You were asked to write a report, of maxim 8 pages
( .... your community members appreciate effective communication)
, addressing issues such as:
what Julie Payette's case says about employee-employer relations in Canada?
what Bill Morneau's case says about ethics in Canada?
what Jody Wilson-Raybould's case says about globalization, global competition, competitiveness and ethics in Canada?
Your community is generally optimistic about the state of affairs in Canada, and about the future of the country which depends on its functioning democracy.
Are there warning signs and "red flags" to watch for by engaged members of the Canadian society?"
.
You are a new university police chief in a medium-sized city, an.docxadampcarr67227
You are a new university police chief in a medium-sized city, and today is a huge football game. You have received information from a patrol sergeant that one of your male officers is at the football stadium working overtime and wearing an earring and sporting a new, visible and rather risqué tattoo on his lower front arm. The sergeant says that both are highly visible, and that a rudimentary dress code exists in your agency but does not cover earrings. You are aware that the other officers are anxiously watching the situation to see what you do. What are you going to do? Explain yourself.
.
You are a native speaker of French living in a mainly English speaki.docxadampcarr67227
You are a native speaker of French living in a mainly English speaking part of Canada. You would like to send your children to a French school, but none is available. Remembering how the Gaulois culture and language progressively disappeared in what is now France, you would like to alert the French speaking population and its leaders to the importance of having a Francophone system of education
400-500 words
double spaced
tiems new roman
I need by nov 19th at 4pm
.
You are a new high school teacher, and have been captured at the end.docxadampcarr67227
You are a new high school teacher, and have been captured at the end of Open House by a parent who is upset about one of your classroom procedures. You have tried to explain the value of the procedure; however, the parent continues to adamantly disagree and hold you hostage after everyone has left. What do you think would be the best course of action?
.
You are a member of the Human Resource Department of a medium-sized .docxadampcarr67227
You are a member of the Human Resource Department of a medium-sized organization that is implementing a new inter-organizational system that will impact employees, customers, and suppliers. Your manager has requested that you work with the system development team to create a communications plan for the project. He would like to meet with you in two hours to review your thoughts on the KEY OBJECTIVES OF THE COMMUNICATIONS PLAN. What should those objectives be?
.
You are a network analyst on the fly-away team for the FBIs cyberse.docxadampcarr67227
You are a network analyst on the fly-away team for the FBI's cybersecurity sector engagement division. You've been deployed several times to financial institutions to examine their networks after cyberattacks, ranging from intrusions and data exfiltration to distributed denial of services to their network supporting customer transaction websites. A representative from the Financial Services Information Sharing and Analysis Center, FS-ISAC, met with your boss, the chief net defense liaison to the financial services sector, about recent reports of intrusions into the networks of banks and their consortium.
He's provided some of the details of the reports in an email. "Millions of files were compromised, and financial officials want to know who entered the networks and what happened to the information. At the same time, the FS-ISAC has seen extensive distributed denial of service disrupting the bank's networks, impacting the customer websites, and blocking millions of dollars of potential transactions," his email reads.
You realize that the impact from these attacks could cause the downfall of many banks and ultimately create a strain on the US economy. In the email, your chief asks you to travel to one of the banks and using your suite of network monitoring and intrusion detection tools, produce two documents—a report to the FBI and FS-ISAC that contains the information you observed on the network and a joint network defense bulletin to all the banks in the FS-ISAC consortium, recommending prevention methods and remediation against the types of malicious traffic activity that they may face or are facing.
Network traffic analysis and monitoring help to distinguish legitimate traffic from malicious traffic. Network administrators must protect networks from intrusions. This can be done using tools and techniques that use past traffic data to determine what should be allowed and what should be blocked. In the face of constantly evolving threats to networks, network administrators must ensure their intrusion detection and prevention systems are able to analyze, monitor, and even prevent these advanced threats.
In this project, you will research network intrusion and prevention systems and understand their use in a network environment. You will also use monitoring and analysis technologies in the Workspace to compile a Malicious Network Activity Report for financial institutions and a Joint Network Defense Bulletin for a financial services consortium.
The following are the deliverables for this project:
Deliverables
•Malicious Network Activity Report: An eight- to 10-page double-spaced Word document with citations in APA format. The page count does not include figures, diagrams, tables, or citations.
•Joint Network Defense Bulletin: A one- to two-page double-spaced document.
Step 1: Create a Network Architecture Overview
You travel to the various bank locations and gain access to their networks. However, yo.
You are a member of the senior management staff at XYZ Corporation. .docxadampcarr67227
You are a member of the senior management staff at XYZ Corporation. You have historically been using a functional structure set up with five departments: finance, human resources, marketing, production, and engineering.
Create a drawing of your simplified functional structure, identifying the five departments.
Assume you have decided to move to a project structure. What might be some of the environmental pressures that would contribute to your belief that it is necessary to alter the structure?
With the project structure, you have four projects currently ongoing: stereo equipment, instrumentation and testing equipment, optical scanners, and defense communications.
Draw the new structure that creates these four projects as part of the organizational chart.
Text
Title:
Project Management
ISBN: 9780134730332
Authors: Pinto
Publisher: Pearson
Edition: 5TH 19
.
You are a member of the senior hospital administration. You become a.docxadampcarr67227
You are a member of the senior hospital administration. You become aware of a problem involving a long-time and well-respected employee, as well as the supervisor of said employee.
The employee in question is a social worker; a very competent and very conscientious professional. His wife has recently suffered a stroke with significant residual neurological deficit. This has resulted in the necessity that the social worker take days off to care for her; come in late or leave early to take her to medical, physical, or occupational therapy appointments; etc.
It is thought that, because of these demands on his time—and the taxing emotional overlay of dealing with the critical illness of a loved one, while simultaneously dealing with patients and families in similar situations—that his charting fell behind. In fact, it was discovered that he was writing social work notes 1–2 days after the fact, back-dating the notes, and placing them in the patients chart between notes of the same time frame as the date on the note.
When the social worker’s immediate supervisor became aware of this, she told him that such behavior must stop immediately. Given the circumstances, however, she opted to take no further action, did not document this in his personnel file, nor did she advise her superiors.
Other members of the staff became aware of this, and someone reported it to the CEO via a “Tell Us About Problems” Dropbox.
You have been assigned to address these multiple issues of ethics, standards of conduct, truth, and fairness. Also describe what concepts of change management theory you would apply in this situation.
Describe your answer in detail, citing references in APA format where appropriate. Your Journal entry should be at least 500 words.
.
YOU ARE A MEMBER OF THE SENIOR HOSPITAL ADMINISTRATI.docxadampcarr67227
YOU ARE A MEMBER OF THE SENIOR
HOSPITAL ADMINISTRATION.
YOU BECOME AWARE OF A PROBLEM
INVOLVING A LONG-TIME AND WELL-
RESPECTED EMPLOYEE, AS WELL AS THE
SUPERVISOR OF SAID EMPLOYEE.
THE EMPLOYEE IN QUESTION IS A SOCIAL
WORKER; A VERY COMPETENT AND VERY
CONSCIENTIOUS PROFESSIONAL. HIS WIFE
HAS RECENTLY SUFFERED A STROKE WITH
SIGNIFICANT RESIDUAL NEUROLOGICAL
DEFICIT.
THIS HAS RESULTED IN THE NECESSITY THAT
THE SOCIAL WORKER TAKE DAYS OFF TO CARE
FOR HER; COME IN LATE OR LEAVE EARLY TO
TAKE HER TO MEDICAL, PHYSICAL, OR
OCCUPATIONAL THERAPY APPOINTMENTS; ETC.
THAT HIS
CHARTING
FELL BEHIND.
IT IS THOUGHT THAT, BECAUSE OF THESE DEMANDS ON HIS
TIME—AND THE TAXING EMOTIONAL OVERLAY OF DEALING
WITH THE CRITICAL ILLNESS OF A LOVED ONE, WHILE
SIMULTANEOUSLY DEALING WITH PATIENTS AND FAMILIES
IN SIMILAR SITUATIONS—
WHEN THE SOCIAL WORKER’S IMMEDIATE
SUPERVISOR BECAME AWARE OF THIS, SHE TOLD.
IN FACT, IT WAS DISCOVERED THAT HE
WAS WRITING SOCIAL WORK NOTES 1-2
DAYS AFTER THE FACT, BACK-DATING THE
NOTES, AND PLACING THEM IN THE
PATIENTS CHART BETWEEN NOTES OF THE
SAME TIME FRAME AS THE DATE ON THE
NOTE.
GIVEN THE CIRCUMSTANCES,
HOWEVER, SHE OPTED TO TAKE NO
FURTHER ACTION, DID NOT
DOCUMENT THIS IN HIS PERSONNEL
FILE, NOR DID SHE ADVISE HER
SUPERIORS.
JOURNAL TOPIC
POST YOUR RESPONSE ON
THE UNIT 7 JOURNAL AREA.
Other members of the staff became aware of
this, and someone reported it to the CEO via a
“Tell Us About Problems” drop box.
You have been assigned to address these
multiple issues of ethics, standards of conduct,
truth, and fairness. Also describe what concepts
of change management theory you would apply
in this situation.
Describe your answer in detail, citing references
in APA format where appropriate. Your Journal
entry should be at least 500 words.
Slide Number 1Slide Number 2Slide Number 3Slide Number 4
.
You are a member of the Human Resource Department of a medium-si.docxadampcarr67227
You are a member of the Human Resource Department of a medium-sized organization that is implementing a new inter organizational system that will impact employees, customers, and suppliers. Your manager has requested that you work with the system development team to create a communications plan for the project. He would like to meet with you in two hours to review your thoughts on the KEY OBJECTIVES OF THE COMMUNICATIONS PLAN. What should those objectives be?
.
You are a member of the American Indian tribe. Think about how your .docxadampcarr67227
You are a member of the American Indian tribe. Think about how your life has changed since the English settlers (Plymouth Colonists) have settled on your land. How do you feel with them there? Are you happy? Are they happy? Write a letter to the colonists expressing your feelings. Bring in historical facts to make your letter believeable.
Your letter should include:
Describe your life before the arrival of the English settlers.
What were your first impressions on the settlers?
How has having the settlers live nearby changed your life?
Do you think the English settlers have the right to settle in Plymouth? Why or why not?
What can the settlers learn form you, and what can you learn from the settlers?
How can two cultures live together peacefully? What would you have to do to make this happen?
.
You are a juvenile justice consultant creating a proposal that w.docxadampcarr67227
You are a juvenile justice consultant creating a proposal that will be presented to the state legislature concerning the future of the juvenile justice system.
Create
a 10- to 15-slide Microsoft® PowerPoint® presentation, including speaker notes, detailing your proposal. Address recommendations for all aspects of the system, including:
Community involvement
Law enforcement
Courts and sentencing
Corrections
Include
a justification for the system based on history, trends, causation theories, and potential for reform.
.
You are a journalist and you have been sent off to write a story abo.docxadampcarr67227
You are a journalist and you have been sent off to write a story about a break in at a local school. You write for the local paper entitled The Local Post. This is the information that you have got so far.
Things that were stolen include:
Five laptop computers
Money that was raised for Comic Relief
Two digital cameras
The school is called Rosedale Primary School and the Head teacher's name is Mr John Jones.
People that could be interviewed are:
The Head teacher
Mrs Milton - a parent
Mr Thompson - lives down the road
The police have investigated and viewed the CCTV footage. There are two men seen committing this crime, covered in black clothing. Police are appealing for witnesses to come forward.
.
You are a juvenile court probation officer. You have a choice of.docxadampcarr67227
You are a juvenile court probation officer. You have a choice of programs including; mandatory counseling, family counseling, removal from the home and placing in foster care, diversion, incarceration in a youth home or mandatory participation in a 10 week boot camp. You must make recommendations to the judge for sentencing. You must use all the alternatives for the group and you can’t use more than one alternative twice. Make recommendations for each juvenile and explain your rationale. Note your difficulties and what further information you would have liked. Finally what is the overwhelming need for each person and how are you addressing that in your program.
Sally is 13 and lives in the suburbs of Fort Wayne. She was caught riding in a stolen car with two friends from high school. Sally has no record – her mother tells you that Sally was a model child until last year when her father died. Since then Sally’s grades have dropped and she has become unmanageable.
John is 16 and lives in Indianapolis. He has a long juvenile record dating back to when he was 10. John’s prior offenses include arson, disorderly conduct, larceny and assault (3). John was arrested for stealing lawn ornaments worth $23.00. John is unsupervised (no parental control) and missed his last probation meeting.
Don is 14 and lives in the inner-city of Gary, Indiana. Don has no father and his mother is a crack addict. Don lives by himself for long periods of time. In the past Don was arrested for stealing food from a local bakery. Don admitted to the theft, but noted he hadn’t eaten in two days. Don was removed from home – but was returned to his mother one year later. Don was arrested for possession of crack cocaine – it was believed he was selling.
Darlene is 12 and lives in the suburbs with her mother, step-father and new baby sister. Darlene has been in juvenile court a number of times in the past year for being a runaway. She was petitioned last month by her step-father for being incorrigible. Darlene refused to follow the family rules and is defiant to her step-father. Darlene is very intelligent and is openly disrespectful to her mother and step-father.
Stephen Holmes is 16 and lives in Noblesville. His father is a salesman and his mother is an executive with General Advertising Inc. Stephen has a prior record for larceny. Last month Stephen got into a fight with his brother who is 17. After the fight was over Stephen took his father’s gun and shot his brother in the head instantly killing him.
Papers will be completed in Word Format as an attachment. The papers will be typed in Times New Roman using 12 font. Papers will be double-spaced. The papers will be at least 500 words in length. The papers will be a critical examination of a topic area chosen by the instructor. Students are encouraged to critically examine and question a topic area in detail using their book.
.
Level 3 NCEA - NZ: A Nation In the Making 1872 - 1900 SML.pptHenry Hollis
The History of NZ 1870-1900.
Making of a Nation.
From the NZ Wars to Liberals,
Richard Seddon, George Grey,
Social Laboratory, New Zealand,
Confiscations, Kotahitanga, Kingitanga, Parliament, Suffrage, Repudiation, Economic Change, Agriculture, Gold Mining, Timber, Flax, Sheep, Dairying,
This presentation was provided by Racquel Jemison, Ph.D., Christina MacLaughlin, Ph.D., and Paulomi Majumder. Ph.D., all of the American Chemical Society, for the second session of NISO's 2024 Training Series "DEIA in the Scholarly Landscape." Session Two: 'Expanding Pathways to Publishing Careers,' was held June 13, 2024.
A Visual Guide to 1 Samuel | A Tale of Two HeartsSteve Thomason
These slides walk through the story of 1 Samuel. Samuel is the last judge of Israel. The people reject God and want a king. Saul is anointed as the first king, but he is not a good king. David, the shepherd boy is anointed and Saul is envious of him. David shows honor while Saul continues to self destruct.
Elevate Your Nonprofit's Online Presence_ A Guide to Effective SEO Strategies...TechSoup
Whether you're new to SEO or looking to refine your existing strategies, this webinar will provide you with actionable insights and practical tips to elevate your nonprofit's online presence.
Temple of Asclepius in Thrace. Excavation resultsKrassimira Luka
The temple and the sanctuary around were dedicated to Asklepios Zmidrenus. This name has been known since 1875 when an inscription dedicated to him was discovered in Rome. The inscription is dated in 227 AD and was left by soldiers originating from the city of Philippopolis (modern Plovdiv).
Philippine Edukasyong Pantahanan at Pangkabuhayan (EPP) CurriculumMJDuyan
(𝐓𝐋𝐄 𝟏𝟎𝟎) (𝐋𝐞𝐬𝐬𝐨𝐧 𝟏)-𝐏𝐫𝐞𝐥𝐢𝐦𝐬
𝐃𝐢𝐬𝐜𝐮𝐬𝐬 𝐭𝐡𝐞 𝐄𝐏𝐏 𝐂𝐮𝐫𝐫𝐢𝐜𝐮𝐥𝐮𝐦 𝐢𝐧 𝐭𝐡𝐞 𝐏𝐡𝐢𝐥𝐢𝐩𝐩𝐢𝐧𝐞𝐬:
- Understand the goals and objectives of the Edukasyong Pantahanan at Pangkabuhayan (EPP) curriculum, recognizing its importance in fostering practical life skills and values among students. Students will also be able to identify the key components and subjects covered, such as agriculture, home economics, industrial arts, and information and communication technology.
𝐄𝐱𝐩𝐥𝐚𝐢𝐧 𝐭𝐡𝐞 𝐍𝐚𝐭𝐮𝐫𝐞 𝐚𝐧𝐝 𝐒𝐜𝐨𝐩𝐞 𝐨𝐟 𝐚𝐧 𝐄𝐧𝐭𝐫𝐞𝐩𝐫𝐞𝐧𝐞𝐮𝐫:
-Define entrepreneurship, distinguishing it from general business activities by emphasizing its focus on innovation, risk-taking, and value creation. Students will describe the characteristics and traits of successful entrepreneurs, including their roles and responsibilities, and discuss the broader economic and social impacts of entrepreneurial activities on both local and global scales.
Hypothesis TestingHypothesisThe formal testing of hypothesis.docx
1. Hypothesis Testing
Hypothesis
The formal testing of hypothesis is a critical step not to be
ignored. One of the jobs of statistics is to reduce the
uncertainty in our decision making. It does so by helping to
identify assumptions and check the validity, in essence, what is
really there. It is my personal preference to evaluate stocks or
mutual funds to see if they actually exceed the average market
return of 10%. The 10% return is the historical return on stocks
over the past 75 years, however, each year has its own unique
return. When doing so, I am interested in did this mutual fund
exceed the 10% annual return. When you look at hypothesis
testing there are three ways you can set them up. There are as
follows:
HO: = H1: not equal Word In story problem: Is
Different
Ho: < or = H1: is greater than Word In story problem:
more than, increased
HO: > or = H1: Is less than Word In story problem:
decreased, dropped
The hypothesis come in “pre-packaged sets” so you can’t mix
and match. So when you are setting them up, look at the
alternative (H1) in setting them up. Many times this is the
decision that interests you the most. In this case, I wanted
returns that exceeded ten percent, so I looked at the alternative
hypothesis of greater than when setting up the process. What is
helpful in thinking about the hypothesis and alternative
hypothesis are the unique qualities of the null or HO,
hypotheses. The HO hypothesis has unique characteristics. The
unique characteristics is that you assume there to either be “no
difference” or status quo. So, I set the hypothesis up as
follows:
2. HO: return is less than or equal to ten percent
H1: return is greater than ten percent
For the hypothesis test, I could have chosen to have it set up
two other ways and those are:
HO: the return is equal to ten percent
H1; the return is not equal to ten percent
HO: The return is greater than or equal to ten percent
H1: The return is less than ten percent.
What is helpful in thinking about the hypothesis and alternative
hypothesis are the unique qualities of the null or HO,
hypotheses. The HO hypothesis has unique characteristics. The
unique characteristics are that you assume there to either “no
difference” or status quo. In the stating of HO there is always
an implied equal. To help you understand this concept lets take
a trip to the local vending machine.
Hypothesis Test Thinking and a Vending Machine
Imagine yourself walking up to a vending machine. You want
to purchase a bottle of Pepsi for a $1.00. You walk up to the
vending machine and assume you are going to put your dollar in
the machine, press the Pepsi button, and the machine will
dispense the bottle of Pepsi. Do you walk up to the machine,
knowing that it is not going to give you your bottle of Pepsi
after you insert your dollar? If so, would you proceed with the
transaction. No you wouldn’t make the transaction and
willingly lose your dollar. You assume that as you approach the
machine and put your money in that you will receive your Pepsi
because that is what has happened time after time, this is the
“status quo” segment of the hypothesis. Now, when the
machine doesn’t give you your Pepsi but keeps your dollar, you
3. REJECT the hypothesis that this machine is fair and ACCEPT
the alternative hypotheses that this machine has stolen my
money and is a crooked machine. Now that you know about the
formal hypothesis, let’s look at the rest of the procedure to
separate fact from fiction.
Remaining Hypothesis Testing Steps
Decision Rule:
The remaining steps are to set up the rules and check our
results. We set our rules up ahead of time so as not to be
unduly influenced by the results. So, we decide on a level of
significance. This is a significant level because it determines
the point at which we either accept or reject our null hypothesis.
The levels are normally.10, .05 or .01. These are similar to
purchasing a Pepsi in a small, medium or large, sorry; we don’t
have a super size in statistics. When choosing the levels of
significance, one can look at industry norm or you can choose
any one of the three. Normally, the .05 level of significance is
chosen because this is middle ground. So we chose the .05
level of significance.
Test Statistic
The next step is to calculate our “test statistic”. This is the
statistic that our computer software generates for us. This is
most easily expressed in probability or what is normally called
the “P” value. The “P” stands for probability and can be
thought of this way. If the “true average”, the average that is
really there, not our sample results, the probability of getting a
sample average which we found, is some value. Lets say that
we were testing the hypothesis that our mutual fund return is
less than or equal to ten percent and we input our data, test the
hypothesis at the .05 level of significance. The computer
software calculates a “P value” of .15. A .15 value is greater
4. than the .05 level of significance and so we don’t reject our
hypothesis and conclude the stock we are looking at has a return
less than or equal to 10%. However, if the “ P value”
calculated from the computer program was .04; we would have
rejected the null hypothesis and concluded that the average
return of this mutual fund was greater than ten percent.
Side Note: if the “P or probability value calculated “is less than
the level of significance, we reject the null hypothesis and
accept the alternative. We do this because we are saying that if
the null hypothesis is true, the probability of getting a sample
average with this result is less than the decision rule statistic.
Decision:
The decision is where you compare the probability value
calculated against the critical probability value set in your
decision rule. If your test probability value is less than the
critical, reject the null hypothesis and accept the alternative
hypothesis. The P value testing allows for you to be slightly
confident, if is just on the border line of accept or reject to very
confident if the probability value is very small such as .001
(1/1000)
You may need to double check your values and if unsure, you
can ask yourself do I want a 5% raise or do I want a .1% raise.
This is a simple way of keeping your decision correct and not
confusing the decision because you misinterpreted the p values
size.
Traditional Approach
In 2004, a small dealership leased 21 Chevrolet Impalas on two
year leases. When the cars were returned in 2006, the mileage
recorded was (see table). Is the dealers mean significantly
greater than the national average of 30,000 miles for two year
5. leased vehicles at the 10% level of significance?
T value = 21-1 = 20 , one tailed test, .10 level of significance
= 1.325
HO: The average miles is less than or equal to 30,000
H1: The average miles are greater than 30,000
Decision Rule: If the T value calculated is greater than 1.325,
reject HO
Test Statistic: T= ( 33,950-30,000)/ (11,866/√21)
= ( 33,950-30,000)/ (11,866/4.58)
= 3,950/ 2,590
T value =1.52
Decision: Since the Test statistic of 1.52 is greater than the T
critical value of 1.325, we reject HO and conclude the cars have
more mileage.
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P value Approach
HO: The average miles is less than or equal to 30,000
H1: The average miles are greater than 30,000
Decision Rule: If the P value calculated is less than .10, reject
HO
Test Statistic: .071
Decision: Since the Test statistic of .071 is less than the P
critical value of .10, we reject HO and have weak evidence to
conclude the cars have more mileage. For if the HO was true
that the cars had less than or equal to 30,000 miles, the
probability that you would have randomly found a sample
average of 33,950 is 7.1%, therefore, it is possible but not
probable.
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Box Plot Explanation: The red HO circle is the null hypothesis
value. The blue line is the confidence interval and the xbar is
the sample average from which a 90% confidence interval was
created. Since the HO value is contained in the confidence
interval, it is possible that it could be the “real mean”.
36. .05 Level of Significance
HO: The average return on the Investment Company of
America is equal to 10%
H1: The average return on the Investment Company of America
is not equal to 10%
Decision Rule: If the Z value is less than -1.96 or greater than
1.96, reject HO
Test Statistic: Z = (14.52 – 10)/(18.668/√71)
Z = 4.52/2.21
Z = 2.04
Decision: Since the the Z value calculated of 2.04 is greater
than the Z critical of 1.96 we reject HO and conclude the fund
annual return is different than 10%
Since this was a two tailed test, the .05 reject region was
divided equally on both sides, thus is gives a .025 rejection
zone on each side. Since 50% probability is on each side of the
bell curve, we calculate the z value by 50% - 2.5% = 47.5%
.4750 = Z value of 1.96
NOTES:
2.21 is the standard error of the mean, when the standard
deviation is adjusted to reflect sample size based upon the
theory of the law of large numbers that as sample size increases,
variation decreases.
38. P
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Suppose we are interested now in does the fund beat the
historical average of 10%. It will take the following steps.
Note: This is a one tailed test and you need to remember which
is the positive side and negative side of the bell curvewith Left
is negative, right is positive
HO: The average return on the Investment Company of
39. America is less than or equal to 10%
H1: The average return on the Investment Company of America
is greater than 10%
Decision Rule: If the Z value is greater than Z 1.65, reject HO
Test Statistic: Z = (14.52 – 10)/(18.668/√71)
Z = 4.52/2.21
Z = 2.04
Decision: Since the the Z value calculated of 2.04 is greater
than the Z critical of 1.65,we reject HO and conclude the fund
annual return is greater than 10%
Since this was a one tailed test, the .05 reject region is located
on one side of the bell curve, thus to gain a Z value we subtract
our level of significance, in this case .05 thus 50% - 5% = 455%
.45 = Z value of 1.645
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Suppose we are interested now in does the fund less the
historical average of 10%. It will take the following steps.
HO: The average return on the Investment Company of
America is greater than or equal to 10%
H1: The average return on the Investment Company of America
is less than 10%
Decision Rule: If the Z value is less than Z -1.65, reject HO
Test Statistic: Z = (14.52 – 10)/(18.668/√71)
Z = 4.52/2.21
Z = 2.04
Decision: Since the the Z value calculated of 2.04 is greater
42. than the Z critical of
- 1.65,we fail reject HO and conclude the fund annual return is
greater than or equal to 10%
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44. The formal steps look like this:
One Mutual Fund Scenario
One-Sample Z: Investment Company of America
Test of mu = 10 vs not = 10
The assumed standard deviation = 18.668
Variable N Mean StDev SE Mean 95% CI
Z
Investment Company of Am 71 14.52 18.67 2.22 (10.18,
18.86) 2.04
Variable P
Investment Company of Am 0.041
HO: The average return on the Investment Company of
America is equal to 10%
H1; The average return of the Investment Company of America
of is not equal to 10%
Decision Rule: If the P value is less than .05 reject the HO
Test Statistic: .041
Decision: Since the probability statistics calculated of .041 is
less than the critical probability value of .05, we reject the null
hypothesis and are somewhat confident that the return on this
fund is not equal to 10%. For if the HO was true that the funds
real average is10%, the probability that you would have
45. randomly found a sample average of 14.51 is 4.1%, therefore, it
is possible but not probable
NOTES
One-Sample Z: Investment Company of America
Test of mu = 10 vs > 10
The assumed standard deviation = 18.668
95% Lower
Variable N Mean StDev SE Mean Bound Z
P
Investment Company of Am 71 14.52 18.67 2.22 10.88
2.04 0.021
Suppose we are interested now in does the fund beat the
historical average of 10%. It will take the following steps.
HO: The average return on the Investment Company of
America is less than or equal to 10%
H1: The average return on the Investment Company of America
is greater than 10%
Decision Rule: If the P value calculate is less than .05, reject
HO
Test Statistic: .021
Decision: Since the probability statistics calculated of .021 is
less than the critical probability value of .05, we reject the null
hypothesis and are somewhat confident that the return on this
fund is greater than 10%. For if the HO was true that the funds
real average is less than or equal to10%, the probability that
46. you would have randomly found a sample average of 14.51 is
2.1%, therefore, it is possible but not probable
NOTES
One-Sample Z: Investment Company of America
Test of mu = 10 vs < 10
The assumed standard deviation = 18.668
95% Upper
Variable N Mean StDev SE Mean Bound Z
P
Investment Company of Am 71 14.52 18.67 2.22 18.16
2.04 0.979
Suppose we are interested now in does the fund earns less than
the historical average of 10%. It will take the following steps.
HO: The average return on the Investment Company of
America is greater than or equal to 10%
H1: The average return on the Investment Company of America
is less than 10%
Decision Rule: If the P value calculate is less than .05, reject
HO
Test Statistic: .979
Decision: Since the probability statistics calculated of .979 is
greater than the critical probability value of .05, we fail to
reject the null hypothesis and are very confident that the return
on this fund is greater than or equal to 10%. For if the HO was
true that the funds real average is greater than or equal 10%, the
47. probability that you would have randomly found a sample
average of 14.51 is 97.9%, therefore, it is possible and
probable.
NOTES:
Hypothesis Testing Notes
E-Tickets
Most air traffic passenger use e-tickets. Electronic ticketing
allows passengers not worry about paper tickets and it reduces
the airlines costs. However in recent times airlines have begun
to receive complaints from passengers regarding e-tickets,
especially when having to switch planes. To investigate this
problem an independent watchdog agency contacted a random
sample of 21 airports and collected data on the number of
complaints due to e-tickets.
At the .05 level of significance is there evidence to conclude
that the ten year average of 15 complaints per month has now
changed and actually increased due to electronic ticketing?
Problem Statement:
HO:
H1
Decision Rule
Test Statistic:
Decision
NOTES
1
6
1
60. E-Tickets
Most air traffic passenger use e-tickets. Electronic ticketing
allows passengers not worry about paper tickets and it reduces
the airlines costs. However in recent times airlines have begun
to receive complaints from passengers regarding e-tickets,
especially when having to switch planes. To investigate this
problem an independent watchdog agency contacted a random
sample of 21 airports and collected data on the number of
complaints due to e-tickets.
At the .05 level of significance is there evidence to conclude
that the ten year average of 15 complaints per month has now
changed and actually increased due to electronic ticketing?
Problem Statement:
The airline industry has undergone significant changes
including the movement to e-tickets to reduce costs of paper
tickets as well as allow for internet booking and reservation.
However, given the tight markets and strong competition, the
airline knows it needs to minimize consumer complaints as this
translates into customer defection and loss of revenue.
T test is one tailed (.05) with sample size 21 – 1 = 20 degrees
of freedom = 1.725
HO: Complaints are less than or equal to 15
H1: Complaints are greater than 15
Decision Rule If T is > 1.725, reject HO
Test Statistic: Test Statistic: T = (13.751 – 15)/(1.502/√21)
T= -1.429/.3277
61. T = -4.36
Decision: Since T calculated of -4.36 is less than T critical of
1.725, we fail to reject HO.
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63. E-Tickets
Most air traffic passenger use e-tickets. Electronic ticketing
allows passengers not worry about paper tickets and it reduces
the airlines costs. However in recent times airlines have begun
to receive complaints from passengers regarding e-tickets,
especially when having to switch planes. To investigate this
problem an independent watchdog agency contacted a random
sample of 21 airports and collected data on the number of
complaints due to e-tickets.
At the .05 level ofsignificance is there evidence to conclude that
the ten year average of 15 complaints per month has now
changed and actually increased due to electronic ticketing?
Problem Statement:
The airline industry has undergone significant changes
including the movement to e-tickets to reduce costs of paper
tickets as well as allow for internet booking and reservation.
However, given the tight markets and strong competition, the
airline knows it needs to minimize consumer complaints as this
translates into customer defection and loss of revenue.
HO: Complaints are less than or equal to 15
H1: Complaints are greater than 15
Decision Rule: If P calculated is less than .05, reject HO
Test Statistic: 1.00 actually is .9999 but was rounded up to one
Decision: Since the probability value calculated of .9999 is
greater than .05, we fail to reject HO and conclude there is
strong evidence to support that the number of complaints has
not increased. For if the HO was true that the funds real average
of complaints are less than or equal to 15, the probability that
you would have randomly found a sample average of 13.571 is
64. 99.9%, therefore, it is possible and very probable.
Test of mu = 15 vs > 15
95% Lower
Variable N Mean StDev SE Mean Bound T P
E-Ticket 21 13.571 1.502 0.328 13.006 -4.36 1.000
Chicken Feed
Raising chickens in commercial chicken farms is a growing
industry, as there is a shift from red meat to more white meat.
New Jersey Red Chickens are a favorite chicken and the Feed is
Us company has a new chicken food that they claim is excellent
for increasing weight and naturally it costs more. You have
heard this line before and therefore are skeptical but the
potential pay off is huge. You buy a small amount and feed it
to ten chickens, which you choose at random.
At the .01 level of significance is there evidence that the
chicken weight exceeds the average chicken weight of 4.35
pounds.
Problem Statement:
HO:
H1
Decision Rule
Test Statistic:
Decision
Notes:
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g
h
t
s
Chicken Feed
Raising chickens in commercial chicken farms is a growing
industry, as there is a shift from red meat to more white meat.
New Jersey Red Chickens are a favorite chicken and the Feed is
Us company has a new chicken food that they claim is excellent
for increasing weight and naturally it costs more. You have
heard this line before and therefore are skeptical but the
potential pay off is huge. You buy a small amount and feed it
to ten chickens, which you choose at random.
At the .01 level of significance is there evidence that the
chicken weight exceeds the average chicken weight of 4.35
pounds.
Problem Statement:
HO: The chicken weight is less than or equal to 4.35 pounds
H1: The chicken weight is greater than 4.35 pounds
Decision Rule If T is > 2.821, reject HO
T test is one tailed (.01) with sample size 10 – 1 = 9 degrees of
freedom = 2.821
Test Statistic: Test Statistic: T = (4.368 – 4.35)/(.0339/√10)
T= .018/.0107
78. T = 1.68
Decision: Since T calculated of 1.68 is less than T critical of
2.821, we do not reject HO.
Notes:
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80. Chicken Feed
Raising chickens in commercial chicken farms is a growing
industry, as there is a shift from red meat to more white meat.
New Jersey Red Chickens are a favorite chicken and the Feed is
Us company has a new chicken food that they claim is excellent
for increasing weight and naturally it costs more. You have
heard this line before and therefore are skeptical but the
potential pay off is huge. You buy a small amount and feed it
to ten chickens, which you choose at random.
At the .01 level of significance is there evidence that the
chicken weight exceeds the average chicken weight of 4.35
pounds.
Problem Statement:
HO: The chicken weight is less than or equal to 4.35 pounds
H1: The chicken weight is greater than 4.35 pounds
Decision Rule: If P calculated is less than .01, reject HO
Test Statistic: .064
Decision: Since the P value calculated of .064 is greater than
.01, we fail to reject HO and are only somewhat confident the
real chicken weight is less than or equal to 4.35 pounds For if
the HO was true that the real average chicken weigh is less than
or equal to 4.35, the probability that you would have randomly
found a sample average of 4.368 is 6.4%, therefore, it is
possible and probable.
Issue: small sample size, law of large numbers as sample size
increases variation decreases.
Test of mu = 4.35 vs > 4.35
Small sample law of large number
99% Lower
Variable N Mean StDev SE Mean Bound T P
84. i
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)
Confidence interval contains the null hypothesis, so it is
possible it contains the real average.
Notes:
AARP Work Survey
The American Association of Retired Persons (AARP) reports
that 60% of retired persons under the age of 65 would be
willing to return to work on a full time basis if a suitable job
were available. A sample of 500 retired persons under the age
of 65 revealed 315 would return to work. At the .05 level of
significance can we conclude that more than 60% of retired
people in the age group would return to work?
85. Problem Statement:
In today’s economy, each person is responsible for their own
retirement investment decisions. Many retirees are forced to
return to work to pay bills since their retirement savings are
inadequate and there are also those who return to work for
socialization and wanting to continue to make contributions to
society. The question is today are more senior citizens
returning to work than in previous generations? This has
profound social and economic ramifications.
HO:
H1
Decision Rule
Test Statistic:
Decision
Notes:
AARP Work Survey
The American Association of Retired Persons (AARP) reports
that 60% of retired persons under the age of 65 would be
willing to return to work on a full time basis if a suitable job
were available. A sample of500 retired persons under the age
of 65 revealed 315 would return to work. At the .05 level of
significance can we conclude that more than60% of retired
people in the age group would return to work?
Problem Statement:
In today’s economy, each person is responsible for their own
retirement investment decisions. Many retirees are forced to
return to work to pay bills since their retirement savings are
inadequate and there are also those who return to work for
socialization and wanting to continue to make contributions to
society. The question is today are more senior citizens
returning to work than in previous generations? This has
profound social and economic ramifications.
86. Z value is used since it is a proportion
HO: The proportion of retired persons returning to work is less
than or equal to 60%
H1: The proportion of retired persons returning to work is
greater than 60%
Decision Rule: If z calculated is greater than 1.645, reject HO
Test Statistic: Z = .63-.60/ (√.63*.37/500)
Z = 1.38
315/500= .63
1-.63 = .37
Decision Since the Z calculated of 1.38 is less than the critical
value of 1.645, we fail to reject HO and conclude the average is
still less than or equal to 60%.
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1
AARP Work Survey
The American Association of Retired Persons (AARP) reports
that 60% of retired persons under the age of 65 would be
willing to return to work on a full time basis if a suitable job
were available. A sample of 500 retired persons under the age
of 65 revealed 315 would return to work. . At the .05 level of
significance can we conclude that more than60% of retired
people in the age group would return to work?
Problem Statement:
In today’s economy, each person is responsible for their own
retirement investment decisions. Many retirees are forced to
return to work to pay bills since their retirement savings are
inadequate and there are also those who return to work for
socialization and wanting to continue to make contributions to
society. The question is today are more senior citizens
returning to work than in previous generations? This has
profound social and economic ramifications.
HO: The proportion of retired persons returning to work is less
than or equal to 60%
H1: The proportion of retired persons returning to work is
greater than 60%
89. Decision Rule: If P calculated is less than .05, reject HO
Test Statistic: .085
Decision: Since the P value calculated of .085 is greater than
.05, we fail to reject HO and are only somewhat confident the
real percentage of senior citizens returning to work is less than
or equal to 60%. For if the HO was true that the number of
people willing to return to work is less than or equal to 60%,
the probability that you would have randomly found a sample
average of 63% is 8.5%, therefore, it is possible and probable
AARP Work Survey
Test of p = 0.6 vs p > 0.6
95%
Lower
Sample X N Sample p Bound Z-Value P-Value
1 315 500 0.630000 0.594485 1.37 0.085
Hypothesis Test Directions for Minitab
Step One Sample Z Hypothesis Test
Begin by clicking on stat, basic stat then on One Sample Z
Step Two for Z Hypothesis Test
1. Click on the column of data to be used: Investment Company
of America
2. Enter the standard deviation value calculated from the
graphical summary of the data which is 18.66
90. 3. Click on perform hypothesis test
4. Enter the value for the test 10
5. Click on the Options Box
Step Three for Z Hypothesis Test
· Click on options
· Enter the alternative hypothesis test which in this example is
greater than
· Enter the confidence interval value that corresponds with the
level of significance in this example it is 95%
Step Four for Z Hypothesis Test
· Click on graphs
· Check the boxplot of data or other graph of your choosing.
· Click OK
Step Five for Z Hypothesis Test
Click OK to run the test
Step Six for Z Hypothesis Test
Results of the test
Results of the test are now showing in the session window and
you can copy and paste them into a graph or word document.
91. Directions for a One Sample T Test
Step 1
Enter your data and give them column headings for ease of use.
……..
Click on the menu and if you have a sample size smaller than 30
click on the one sample T or if your standard deviation is
calculated from your sample data, you will use a one sample T
test. One sample means you are only comparing one average
calculated from the of data against the historical or status quo
mean.
Step 2
Now that you have chosen the test, you will need to indicate
where the data is found by clicking on the column where the
data is located, unless you have what is called summarized data.
Summarized data is when you have already calculated or been
given the sample size, sample average and sample standard
deviation.
If your data is in a column such as the example, click the data
column heading to move it into the box.
If summarized data, where the mean, standard deviation and
sample size are given, then click on the “summarized data” dot
and enter the information into the boxes.
Step 3
92. · In this step you will need to check the hypothesis test box and
then
· Enter the value tested in this case it is 4.35 pounds.
· Click Options
Step 4
· Enter the confidence interval level that corresponds with the
level of significance. In this case our level of significance is
.05, so the corresponding level of significance is 95%
· Select the correct H1
· Click on OK
Step 5
· In this step you will choose the graph options that would be
most helpful to you. You can check all three and then look at
the options as they appear to decide which is most helpful. In
this case we choose a boxplot
· Click on boxplot
Step 6
Session Window with Results
· The hypothesis test listed with the alternative hypothesis
being referred to as the “vs not”.
93. · The P value associated with the hypothesis test and other
values are listed in the “”Variable output” section of this view.
Step 7
Graphical Analysis
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In graphical analysis you are looking to see if the HO is
contained in the confidence interval as if the confidence interval
is contained, you do not reject HO because it is possible that it
could be the “real average”.
Summarized Data
Step One ( if given, skip this step and go to step two)
109. r
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Prior to using summarized data you can find the information in
a graphical summary but most of the time it is given.
Step Two
Enter the sample size, mean and standard deviation along with
the hypothesized value.
Step Three
· Enter the confidence interval that corresponds with the level
110. of significance
· Enter the H1
Click ok, since our data was summarized, we can’t do graph as
we lack the individual data points.
E-ticket
Chicken Weights
Year ICA
Investment Company of America
14
4.41
2004
19.8
14
4.37
2003
26.3
16
4.33
2002
-14.47
12
4.35
2001
-4.05
12
4.30
2000
121. Positive side
Key Data
HO value status quo
Standard deviation and sample size
Sample average
Fail to Reject
Test Stat 2.04
< - 1.96 Reject
Reject >1.96
122. HO value status quo
Standard deviation and sample size
Sample average
Fail to Reject
Test Stat 2.04
Reject >1.645
HO value status quo
Standard deviation and sample size
Sample average
123. Fail to Reject
Test Stat 2.04
Reject < -1.645
Lower Bound
P value
H1
Sample average
Reject >1.725
Fail to Reject
124. Test Stat -4.36
H1
Sample average
Test Stat 1.68
>2.821 Reject
Fail to Reject
Lower end of confidence interval
Sample average
Null
125. Test Stat 1.38
Fail to Reject
>1.645 Reject
H1
Stat
basic stat
Click on one sample Z
HO value
Standard deviation from graphical summary
126. Column of data
H1
Level of significance
Graph choice
Click OK
One sample T
Check circle if summarized data
Summarized data entry
Click on correct column of data
127. Options
HO value
Check box
Choose the H1
Enter confidence interval level
Graph options
Alternative Hypothesis listed
P value calculated
3rd quartile
129. Choose H1
Choose level of significance
_1296299599
_1296478540
_1296485041
_1296485042
_1296485043
_1296299021
_1266419144
_1266418152
_1266409737
_1265981359
Hypothesis Testing
Hypothesis
The formal testing of hypothesis is a critical step not to be
ignored. One of the jobs of statistics is to reduce the
uncertainty in our decision making. It does so by helping to
identify assumptions and check the validity, in essence, what is
really there. It is my personal preference to evaluate stocks or
mutual funds to see if they actually exceed the average market
return of 10%. The 10% return is the historical return on stocks
130. over the past 75 years, however, each year has its own unique
return. When doing so, I am interested in did this mutual fund
exceed the 10% annual return. When you look at hypothesis
testing there are three ways you can set them up. There are as
follows:
HO: = H1: not equal Word In story problem: Is
Different
Ho: < or = H1: is greater than Word In story problem:
more than, increased
HO: > or = H1: Is less than Word In story problem:
decreased, dropped
The hypothesis come in “pre-packaged sets” so you can’t mix
and match. So when you are setting them up, look at the
alternative (H1) in setting them up. Many times this is the
decision that interests you the most. In this case, I wanted
returns that exceeded ten percent, so I looked at the alternative
hypothesis of greater than when setting up the process. What is
helpful in thinking about the hypothesis and alternative
hypothesis are the unique qualities of the null or HO,
hypotheses. The HO hypothesis has unique characteristics. The
unique characteristics is that you assume there to either be “no
difference” or status quo. So, I set the hypothesis up as
follows:
HO: return is less than or equal to ten percent
H1: return is greater than ten percent
For the hypothesis test, I could have chosen to have it set up
two other ways and those are:
HO: the return is equal to ten percent
131. H1; the return is not equal to ten percent
HO: The return is greater than or equal to ten percent
H1: The return is less than ten percent.
What is helpful in thinking about the hypothesis and alternative
hypothesis are the unique qualities of the null or HO,
hypotheses. The HO hypothesis has unique characteristics. The
unique characteristics are that you assume there to either “no
difference” or status quo. In the stating of HO there is always
an implied equal. To help you understand this concept lets take
a trip to the local vending machine.
Hypothesis Test Thinking and a Vending Machine
Imagine yourself walking up to a vending machine. You want
to purchase a bottle of Pepsi for a $1.00. You walk up to the
vending machine and assume you are going to put your dollar in
the machine, press the Pepsi button, and the machine will
dispense the bottle of Pepsi. Do you walk up to the machine,
knowing that it is not going to give you your bottle of Pepsi
after you insert your dollar? If so, would you proceed with the
transaction. No you wouldn’t make the transaction and
willingly lose your dollar. You assume that as you approach the
machine and put your money in that you will receive your Pepsi
because that is what has happened time after time, this is the
“status quo” segment of the hypothesis. Now, when the
machine doesn’t give you your Pepsi but keeps your dollar, you
REJECT the hypothesis that this machine is fair and ACCEPT
the alternative hypotheses that this machine has stolen my
money and is a crooked machine. Now that you know about the
formal hypothesis, let’s look at the rest of the procedure to
separate fact from fiction.
132. Remaining Hypothesis Testing Steps
Decision Rule:
The remaining steps are to set up the rules and check our
results. We set our rules up ahead of time so as not to be
unduly influenced by the results. So, we decide on a level of
significance. This is a significant level because it determines
the point at which we either accept or reject our null hypothesis.
The levels are normally.10, .05 or .01. These are similar to
purchasing a Pepsi in a small, medium or large, sorry; we don’t
have a super size in statistics. When choosing the levels of
significance, one can look at industry norm or you can choose
any one of the three. Normally, the .05 level of significance is
chosen because this is middle ground. So we chose the .05
level of significance.
Test Statistic
The next step is to calculate our “test statistic”. This is the
statistic that our computer software generates for us. This is
most easily expressed in probability or what is normally called
the “P” value. The “P” stands for probability and can be
thought of this way. If the “true average”, the average that is
really there, not our sample results, the probability of getting a
133. sample average which we found, is some value. Lets say that
we were testing the hypothesis that our mutual fund return is
less than or equal to ten percent and we input our data, test the
hypothesis at the .05 level of significance. The computer
software calculates a “P value” of .15. A .15 value is greater
than the .05 level of significance and so we don’t reject our
hypothesis and conclude the stock we are looking at has a return
less than or equal to 10%. However, if the “ P value”
calculated from the computer program was .04; we would have
rejected the null hypothesis and concluded that the average
return of this mutual fund was greater than ten percent.
Side Note: if the “P or probability value calculated “is less than
the level of significance, we reject the null hypothesis and
accept the alternative. We do this because we are saying that if
the null hypothesis is true, the probability of getting a sample
average with this result is less than the decision rule statistic.
Decision:
The decision is where you compare the probability value
calculated against the critical probability value set in your
decision rule. If your test probability value is less than the
critical, reject the null hypothesis and accept the alternative
hypothesis. The P value testing allows for you to be slightly
confident, if is just on the border line of accept or reject to very
confident if the probability value is very small such as .001
(1/1000)
You may need to double check your values and if unsure, you
can ask yourself do I want a 5% raise or do I want a .1% raise.
This is a simple way of keeping your decision correct and not
confusing the decision because you misinterpreted the p values
size.
134. Traditional Approach
In 2004, a small dealership leased 21 Chevrolet Impalas on two
year leases. When the cars were returned in 2006, the mileage
recorded was (see table). Is the dealers mean significantly
greater than the national average of 30,000 miles for two year
leased vehicles at the 10% level of significance?
T value = 21-1 = 20 , one tailed test, .10 level of significance
= 1.325
HO: The average miles is less than or equal to 30,000
H1: The average miles are greater than 30,000
Decision Rule: If the T value calculated is greater than 1.325,
reject HO
Test Statistic: T= ( 33,950-30,000)/ (11,866/√21)
= ( 33,950-30,000)/ (11,866/4.58)
= 3,950/ 2,590
T value =1.52
Decision: Since the Test statistic of 1.52 is greater than the T
critical value of 1.325, we reject HO and conclude the cars have
more mileage.
135. (
Fail to Reject
) (
Test Stat 1.52
) (
Reject > 1.325
)
(
Key Pieces for Formula
)
P value Approach
HO: The average miles is less than or equal to 30,000
H1: The average miles are greater than 30,000
Decision Rule: If the P value calculated is less than .10, reject
HO
Test Statistic: .071
Decision: Since the Test statistic of .071 is less than the P
critical value of .10, we reject HO and have weak evidence to
conclude the cars have more mileage. For if the HO was true
that the cars had less than or equal to 30,000 miles, the
probability that you would have randomly found a sample
average of 33,950 is 7.1%, therefore, it is possible but not
probable.
136. Box Plot Explanation: The red HO circle is the null hypothesis
value. The blue line is the confidence interval and the xbar is
the sample average from which a 90% confidence interval was
created. Since the HO value is contained in the confidence
interval, it is possible that it could be the “real mean”.
Data
40,060
24,960
14,310
17,370
44,740
44,550
20,250
33,380
24,270
41,740
58,630
35,830
25,750
28,910
25,090
43,380
23,940
43,510
53,680
31,810
36,780
Traditional Method
Investment Company of America
137. (
Negative
side
) (
Positive side
) (
Key Data
)
One Mutual Fund Scenario
.05 Level of Significance
HO: The average return on the Investment Company of
America is equal to 10%
H1: The average return on the Investment Company of America
is not equal to 10%
Decision Rule: If the Z value is less than -1.96 or greater than
1.96, reject HO
(
HO value status quo
)
(
Standard deviation
and sample size
)
(
Sample average
)Test Statistic: Z = (14.52 – 10)/(18.668/√71)
138. Z = 4.52/2.21
Z = 2.04
Decision: Since the the Z value calculated of 2.04 is greater
than the Z critical of 1.96 we reject HO and conclude the fund
annual return is different than 10%
Since this was a two tailed test, the .05 reject region was
divided equally on both sides, thus is gives a .025 rejection
zone on each side. Since 50% probability is on each side of the
bell curve, we calculate the z value by 50% - 2.5% = 47.5%
.4750 = Z value of 1.96
NOTES:
2.21 is the standard error of the mean, when the standard
deviation is adjusted to reflect sample size based upon the
theory of the law of large numbers that as sample size increases,
variation decreases.
(
Fail to Reject
) (
Test Stat 2.04
139. ) (
< - 1.96 Reject
) (
Reject >1.96
)
Suppose we are interested now in does the fund beat the
historical average of 10%. It will take the following steps.
Note: This is a one tailed test and you need to remember which
is the positive side and negative side of the bell curvewith Left
is negative, right is positive
HO: The average return on the Investment Company of
America is less than or equal to 10%
H1: The average return on the Investment Company of America
is greater than 10%
Decision Rule: If the Z value is greater than Z 1.65, reject HO
(
HO value status quo
)
140. (
Standard deviation and sample size
)
(
Sample average
)Test Statistic: Z = (14.52 – 10)/(18.668/√71)
Z = 4.52/2.21
Z = 2.04
Decision: Since the the Z value calculated of 2.04 is greater
than the Z critical of 1.65,we reject HO and conclude the fund
annual return is greater than 10%
Since this was a one tailed test, the .05 reject region is located
on one side of the bell curve, thus to gain a Z value we subtract
our level of significance, in this case .05 thus 50% - 5% = 455%
.45 = Z value of 1.645
(
Fail to Reject
) (
Test Stat 2.04
) (
Reject >
1.645
)
141. Suppose we are interested now in does the fund less the
historical average of 10%. It will take the following steps.
HO: The average return on the Investment Company of
America is greater than or equal to 10%
H1: The average return on the Investment Company of America
is less than 10%
142. Decision Rule: If the Z value is less than Z -1.65, reject HO
(
HO value status quo
)
(
Standard deviation and sample size
)
(
Sample average
)Test Statistic: Z = (14.52 – 10)/(18.668/√71)
Z = 4.52/2.21
Z = 2.04
Decision: Since the the Z value calculated of 2.04 is greater
than the Z critical of
- 1.65,we fail reject HO and conclude the fund annual return is
greater than or equal to 10%
(
Fail to Reject
) (
Test Stat 2.04
) (
143. Reject < -1.645
)
P Value Approach
The formal steps look like this:
One Mutual Fund Scenario
(
Lower Bound
)One-Sample Z: Investment Company of America
Test of mu = 10 vs not = 10
The assumed standard deviation = 18.668
Variable N Mean StDev SE Mean 95% CI
Z
Investment Company of Am 71 14.52 18.67 2.22 (10.18,
18.86) 2.04
(
P value
)
Variable P
Investment Company of Am 0.041
144. HO: The average return on the Investment Company of
America is equal to 10%
H1; The average return of the Investment Company of America
of is not equal to 10%
Decision Rule: If the P value is less than .05 reject the HO
Test Statistic: .041
Decision: Since the probability statistics calculated of .041 is
less than the critical probability value of .05, we reject the null
hypothesis and are somewhat confident that the return on this
fund is not equal to 10%. For if the HO was true that the funds
real average is10%, the probability that you would have
randomly found a sample average of 14.51 is 4.1%, therefore, it
is possible but not probable
NOTES
145. One-Sample Z: Investment Company of America
Test of mu = 10 vs > 10
(
H1
)The assumed standard deviation = 18.668
95% Lower
Variable N Mean StDev SE Mean Bound Z
P
Investment Company of Am 71 14.52 18.67 2.22 10.88
2.04 0.021
Suppose we are interested now in does the fund beat the
historical average of 10%. It will take the following steps.
HO: The average return on the Investment Company of
America is less than or equal to 10%
H1: The average return on the Investment Company of America
is greater than 10%
Decision Rule: If the P value calculate is less than .05, reject
HO
Test Statistic: .021
Decision: Since the probability statistics calculated of .021 is
less than the critical probability value of .05, we reject the null
hypothesis and are somewhat confident that the return on this
fund is greater than 10%. For if the HO was true that the funds
real average is less than or equal to10%, the probability that
146. you would have randomly found a sample average of 14.51 is
2.1%, therefore, it is possible but not probable
NOTES
One-Sample Z: Investment Company of America
Test of mu = 10 vs < 10
The assumed standard deviation = 18.668
95% Upper
Variable N Mean StDev SE Mean Bound Z
P
Investment Company of Am 71 14.52 18.67 2.22 18.16
2.04 0.979
Suppose we are interested now in does the fund earns less than
the historical average of 10%. It will take the following steps.
HO: The average return on the Investment Company of
America is greater than or equal to 10%
H1: The average return on the Investment Company of America
is less than 10%
Decision Rule: If the P value calculate is less than .05, reject
HO
Test Statistic: .979
147. Decision: Since the probability statistics calculated of .979 is
greater than the critical probability value of .05, we fail to
reject the null hypothesis and are very confident that the return
on this fund is greater than or equal to 10%. For if the HO was
true that the funds real average is greater than or equal 10%, the
probability that you would have randomly found a sample
average of 14.51 is 97.9%, therefore, it is possible and
probable.
NOTES:
Hypothesis Testing Notes
E-Tickets
Most air traffic passenger use e-tickets. Electronic ticketing
allows passengers not worry about paper tickets and it reduces
the airlines costs. However in recent times airlines have begun
to receive complaints from passengers regarding e-tickets,
especially when having to switch planes. To investigate this
problem an independent watchdog agency contacted a random
sample of 21 airports and collected data on the number of
complaints due to e-tickets.
At the .05 level of significance is there evidence to conclude
that the ten year average of 15 complaints per month has now
changed and actually increased due to electronic ticketing?
Problem Statement:
HO:
148. H1
Decision Rule
Test Statistic:
Decision
NOTES
E-Tickets
Most air traffic passenger use e-tickets. Electronic ticketing
allows passengers not worry about paper tickets and it reduces
the airlines costs. However in recent times airlines have begun
to receive complaints from passengers regarding e-tickets,
especially when having to switch planes. To investigate this
problem an independent watchdog agency contacted a random
sample of 21 airports and collected data on the number of
complaints due to e-tickets.
At the .05 level of significance is there evidence to conclude
that the ten year average of 15 complaints per month has now
changed and actually increased due to electronic ticketing?
Problem Statement:
The airline industry has undergone significant changes
including the movement to e-tickets to reduce costs of paper
tickets as well as allow for internet booking and reservation.
However, given the tight markets and strong competition, the
airline knows it needs to minimize consumer complaints as this
translates into customer defection and loss of revenue.
149. T test is one tailed (.05) with sample size 21 – 1 = 20 degrees
of freedom = 1.725
HO: Complaints are less than or equal to 15
H1: Complaints are greater than 15
Decision Rule If T is > 1.725, reject HO
(
Sample average
)Test Statistic: Test Statistic: T = (13.751 – 15)/(1.502/√21)
T= -1.429/.3277
T = -4.36
Decision: Since T calculated of -4.36 is less than T critical of
1.725, we fail to reject HO.
(
Reject >1.725
) (
Fail to Reject
) (
Test Stat -4.36
150. )
E-Tickets
Most air traffic passenger use e-tickets. Electronic ticketing
allows passengers not worry about paper tickets and it reduces
the airlines costs. However in recent times airlines have begun
to receive complaints from passengers regarding e-tickets,
especially when having to switch planes. To investigate this
problem an independent watchdog agency contacted a random
sample of 21 airports and collected data on the number of
complaints due to e-tickets.
At the .05 level ofsignificance is there evidence to conclude that
the ten year average of 15 complaints per month has now
changed and actually increased due to electronic ticketing?
Problem Statement:
The airline industry has undergone significant changes
including the movement to e-tickets to reduce costs of paper
tickets as well as allow for internet booking and reservation.
However, given the tight markets and strong competition, the
airline knows it needs to minimize consumer complaints as this
translates into customer defection and loss of revenue.
HO: Complaints are less than or equal to 15
H1: Complaints are greater than 15
Decision Rule: If P calculated is less than .05, reject HO
Test Statistic: 1.00 actually is .9999 but was rounded up to one
151. Decision: Since the probability value calculated of .9999 is
greater than .05, we fail to reject HO and conclude there is
strong evidence to support that the number of complaints has
not increased. For if the HO was true that the funds real average
of complaints are less than or equal to 15, the probability that
you would have randomly found a sample average of 13.571 is
99.9%, therefore, it is possible and very probable.
(
H1
)Test of mu = 15 vs > 15
95% Lower
Variable N Mean StDev SE Mean Bound T P
E-Ticket 21 13.571 1.502 0.328 13.006 -4.36 1.000
Chicken Feed
Raising chickens in commercial chicken farms is a growing
industry, as there is a shift from red meat to more white meat.
New Jersey Red Chickens are a favorite chicken and the Feed is
Us company has a new chicken food that they claim is excellent
for increasing weight and naturally it costs more. You have
heard this line before and therefore are skeptical but the
potential pay off is huge. You buy a small amount and feed it
to ten chickens, which you choose at random.
At the .01 level of significance is there evidence that the
chicken weight exceeds the average chicken weight of 4.35
pounds.
152. Problem Statement:
HO:
H1
Decision Rule
Test Statistic:
Decision
Notes:
Chicken Feed
Raising chickens in commercial chicken farms is a growing
industry, as there is a shift from red meat to more white meat.
New Jersey Red Chickens are a favorite chicken and the Feed is
Us company has a new chicken food that they claim is excellent
for increasing weight and naturally it costs more. You have
heard this line before and therefore are skeptical but the
potential pay off is huge. You buy a small amount and feed it
to ten chickens, which you choose at random.
At the .01 level of significance is there evidence that the
chicken weight exceeds the average chicken weight of 4.35
pounds.
153. Problem Statement:
HO: The chicken weight is less than or equal to 4.35 pounds
H1: The chicken weight is greater than 4.35 pounds
Decision Rule If T is > 2.821, reject HO
T test is one tailed (.01) with sample size 10 – 1 = 9 degrees of
freedom = 2.821
(
Sample average
)Test Statistic: Test Statistic: T = (4.368 – 4.35)/(.0339/√10)
T= .018/.0107
T = 1.68
Decision: Since T calculated of 1.68 is less than T critical of
2.821, we do not reject HO.
Notes:
(
Test Stat 1.68
) (
>2.821 Reject
154. ) (
Fail to Reject
)
Chicken Feed
Raising chickens in commercial chicken farms is a growing
industry, as there is a shift from red meat to more white meat.
New Jersey Red Chickens are a favorite chicken and the Feed is
Us company has a new chicken food that they claim is excellent
for increasing weight and naturally it costs more. You have
heard this line before and therefore are skeptical but the
potential pay off is huge. You buy a small amount and feed it
to ten chickens, which you choose at random.
At the .01 level of significance is there evidence that the
chicken weight exceeds the average chicken weight of 4.35
pounds.
Problem Statement:
HO: The chicken weight is less than or equal to 4.35 pounds
H1: The chicken weight is greater than 4.35 pounds
Decision Rule: If P calculated is less than .01, reject HO
Test Statistic: .064
155. Decision: Since the P value calculated of .064 is greater than
.01, we fail to reject HO and are only somewhat confident the
real chicken weight is less than or equal to 4.35 pounds For if
the HO was true that the real average chicken weigh is less than
or equal to 4.35, the probability that you would have randomly
found a sample average of 4.368 is 6.4%, therefore, it is
possible and probable.
Issue: small sample size, law of large numbers as sample size
increases variation decreases.
Test of mu = 4.35 vs > 4.35
Small sample law of large number
99% Lower
Variable N Mean StDev SE Mean Bound T P
Chicken Weights 10 4.3680 0.0339 0.0107 4.3377 1.68
0.064
(
Lower end of confidence interval
) (
Sample average
) (
Null
)
Confidence interval contains the null hypothesis, so it is
possible it contains the real average.
Notes:
156. AARP Work Survey
The American Association of Retired Persons (AARP) reports
that 60% of retired persons under the age of 65 would be
willing to return to work on a full time basis if a suitable job
were available. A sample of 500 retired persons under the age
of 65 revealed 315 would return to work. At the .05 level of
significance can we conclude that more than 60% of retired
people in the age group would return to work?
Problem Statement:
In today’s economy, each person is responsible for their own
retirement investment decisions. Many retirees are forced to
return to work to pay bills since their retirement savings are
inadequate and there are also those who return to work for
socialization and wanting to continue to make contributions to
society. The question is today are more senior citizens
returning to work than in previous generations? This has
profound social and economic ramifications.
HO:
H1
Decision Rule
Test Statistic:
Decision
157. Notes:
AARP Work Survey
The American Association of Retired Persons (AARP) reports
that 60% of retired persons under the age of 65 would be
willing to return to work on a full time basis if a suitable job
were available. A sample of500 retired persons under the age
of 65 revealed 315 would return to work. At the .05 level of
significance can we conclude that more than60% of retired
people in the age group would return to work?
Problem Statement:
In today’s economy, each person is responsible for their own
retirement investment decisions. Many retirees are forced to
return to work to pay bills since their retirement savings are
inadequate and there are also those who return to work for
socialization and wanting to continue to make contributions to
society. The question is today are more senior citizens
returning to work than in previous generations? This has
profound social and economic ramifications.
Z value is used since it is a proportion
HO: The proportion of retired persons returning to work is less
158. than or equal to 60%
H1: The proportion of retired persons returning to work is
greater than 60%
Decision Rule: If z calculated is greater than 1.645, reject HO
Test Statistic: Z = .63-.60/ (√.63*.37/500)
Z = 1.38
315/500= .63
1-.63 = .37
Decision Since the Z calculated of 1.38 is less than the critical
value of 1.645, we fail to reject HO and conclude the average is
still less than or equal to 60%.
(
Test Stat 1.38
) (
Fail to Reject
) (
>1.645 Reject
)
AARP Work Survey
159. The American Association of Retired Persons (AARP) reports
that 60% of retired persons under the age of 65 would be
willing to return to work on a full time basis if a suitable job
were available. A sample of 500 retired persons under the age
of 65 revealed 315 would return to work. . At the .05 level of
significance can we conclude that more than60% of retired
people in the age group would return to work?
Problem Statement:
In today’s economy, each person is responsible for their own
retirement investment decisions. Many retirees are forced to
return to work to pay bills since their retirement savings are
inadequate and there are also those who return to work for
socialization and wanting to continue to make contributions to
society. The question is today are more senior citizens
returning to work than in previous generations? This has
profound social and economic ramifications.
HO: The proportion of retired persons returning to work is less
than or equal to 60%
H1: The proportion of retired persons returning to work is
greater than 60%
Decision Rule: If P calculated is less than .05, reject HO
Test Statistic: .085
Decision: Since the P value calculated of .085 is greater than
.05, we fail to reject HO and are only somewhat confident the
real percentage of senior citizens returning to work is less than
or equal to 60%. For if the HO was true that the number of
people willing to return to work is less than or equal to 60%,
the probability that you would have randomly found a sample
average of 63% is 8.5%, therefore, it is possible and probable
160. AARP Work Survey
(
H1
)Test of p = 0.6 vs p > 0.6
95%
Lower
Sample X N Sample p Bound Z-Value P-Value
1 315 500 0.630000 0.594485 1.37 0.085
Hypothesis Test Directions for Minitab
(
Stat
basic stat
)Step One Sample Z Hypothesis Test
(
Click on one sample Z
)
1. Begin by clicking on stat, basic stat then on One Sample Z
Step Two for Z Hypothesis Test
(
HO value
) (
Standard deviation from graphical summary
) (
Column of data
)
161. 1. Click on the column of data to be used: Investment Company
of America
2. Enter the standard deviation value calculated from the
graphical summary of the data which is 18.66
3. Click on perform hypothesis test
4. Enter the value for the test 10
5. Click on the Options Box
Step Three for Z Hypothesis Test
(
H1
) (
Level of significance
)
· Click on options
· Enter the alternative hypothesis test which in this example is
greater than
· Enter the confidence interval value that corresponds with the
level of significance in this example it is 95%
Step Four for Z Hypothesis Test
(
Graph choice
)
· Click on graphs
· Check the boxplot of data or other graph of your choosing.
· Click OK
Step Five for Z Hypothesis Test
(
Click OK
)
162. Click OK to run the test
Step Six for Z Hypothesis Test
Results of the test
Results of the test are now showing in the session window and
you can copy and paste them into a graph or word document.
163. Directions for a One Sample T Test
Step 1
Enter your data and give them column headings for ease of use.
(
One sample T
) ……..
Click on the menu and if you have a sample size smaller than 30
click on the one sample T or if your standard deviation is
calculated from your sample data, you will use a one sample T
test. One sample means you are only comparing one average
calculated from the of data against the historical or status quo
mean.
Step 2
(
Click on correct
column
of data
)Now that you have chosen the test, you will need to indicate
where the data is found by clicking on the column where the
data is located, unless you have what is called summarized data.
Summarized data is when you have already calculated or been
given the sample size, sample average and sample standard
deviation.
(
Data in columns
164. ) (
Summarized data entry
)
If your data is in a column such as the example, click the data
column option.
If summarized data, where the mean, standard deviation and
sample size are given, then click on the “summarized data”
option.
Step 3
(
Click on perform hypothesis test box
) (
H
O
value
) (
Click on Chicken Weights
)
· In this step you will need to check the hypothesis test box and
then
· Enter the value tested in this case it is 4.35 pounds.
· Click Options
Step 4
(
Select H1 from list
) (
Enter confidence interval level
165. )
· Enter the confidence interval level that corresponds with the
level of significance. In this case our level of significance is
.05, so the corresponding level of significance is 95%
· Select the correct H1 which is greater than
· Click on OK
Step 5
(
Click OK
) (
Graph options
)
· In this step you will choose the graph options that would be
most helpful to you. You can check all three and then look at
the options as they appear to decide which is most helpful.
· Click OK to run test
Step 6
Session Window with Results
(
P value calculated
) (
Alternative Hypothesis listed
)
· The hypothesis test listed with the alternative hypothesis
166. being referred to as the “vs not”.
· The P value associated with the hypothesis test and other
values are listed in the “”Variable output” section of this view.
Summarized Data
Step One ( if given, skip this step and go to step two)
(
Sample size
) (
Mean, standard deviation
)Prior to using summarized data you can find the information in
a graphical summary but most of the time it is given.
E-ticket
Chicken Weights
Year ICA
Investment Company of America
14
4.41
2004
19.8
14
4.37
2003
26.3
16
4.33
2002
-14.47
12
256. m
m
a
r
y
f
o
r
M
i
l
e
a
g
e
.
Comparing Store Sales
Two Means
Many times we need to compare results to see if our changes in
marketing plans, store changes, changes in product line and
other items to understand if the differences we are seeing are
due to random fluctuation or are they most likely the result of
our influence. Let’s say for example that you are running a
Best Buy electronics store. You have come on board about five
months ago and felt you needed to change your product mix in
your television department by adding flat screens television
sets.
Your secondary research you have conducted over the past three
257. months has yielded some very interesting results concerning flat
panel televisions. One of the studies conducted by Mark
Edwards, a professor at the Stanford School of Business which
focused on lifestyles of individuals with a bachelor’s degree and
earn between 40,000 - $55,000 a year. According to the
research, this target market, plan on replacing their traditional
television sets with flat screen televisions. After reading this
study, you contacted the Chamber of Commerce and located the
demographic information of individuals in the city and found
that you have a very large population who fit the demographics
mentioned in the study by the Stanford Professor. You have
become confident that a change in the marketing mix may boost
sales and profitability. However, before you make a large
change inventory, you add a few more flat panel televisions and
have placed them in the front page of your regular weekly
advertisement.
You have decided to conduct a pilot study to see if there is a
possibility that this change of marketing mix could positively
affect your stores profits. You are doing a pilot study with
limited increases in the breadth and depth of your inventory of
flat screen televisions. This will limit the amount of capital you
need to commit to additional inventory and placed the
promotion in regular advertisement as a means of limiting
additional advertising expenditure and also simulates how you
would promote the product, should the results indicate a full
study be implemented.
Your results seem to indicate there is a positive change in the
sales of televisions. You first conduct a graphical analysis
which indicates there is positive change in televisions and there
appear to be no outlier points influencing the sample averages.
Graphical Analysis
258. Minitab
1400019000240002900034000
95% Confidence Interval for Mu
220002300024000250002600027000280002900030000
95% Confidence Interval for Median
Variable: May Sales
A-Squared:
P-Value:
Mean
StDev
Variance
Skewness
Kurtosis
N
Minimum
1st Quartile
Median
3rd Quartile
Maximum
22586.3
4403.2
22273.4
0.375
0.382
25206.1
5755.4
33124870
-3.6E-01
-3.6E-01
21
13243.0
21728.5
259. 24567.0
30392.5
33456.0
27825.9
8311.2
29242.2
Anderson-Darling Normality Test
95% Confidence Interval for Mu
95% Confidence Interval for Sigma
95% Confidence Interval for Median
Descriptive Statistics
10000200003000040000
95% Confidence Interval for Mu
250002600027000280002900030000310003200033000
95% Confidence Interval for Median
Variable: June Sales
A-Squared:
P-Value:
Mean
StDev
Variance
Skewness
Kurtosis
N
Minimum
1st Quartile
Median
3rd Quartile
Maximum
25897.5
5582.3
25434.0
0.348
261. 123456789101112131415161718192021
Sales Each week
Dollar Value
Sales in MaySales in June
Hypothesis Test
Hypothesis
HO; the mean of the sales for June are less than or equal to May
sales
H1: The mean for June are greater than the mean sale for May.
Decision Rule: If the probability value calculated based upon
the sample is less than the significance level of .05, I will reject
the null hypothesis and conclude the sales of the two months are
equal
P Value = 0.036
Decision: Since the probability value of .036 is less than .05, I
will reject the null hypothesis and conclude the sales in June are
higher than May. However, since the sales figures for only one
month, they would not constitute a trend because of the limited
time frame of the data. However, they would seem to indicate
that longer test marketing would be justified because the
probability value of the test sample of .036 is very low and
therefore the probability that these results are due to just
random variation is very small. If this figure was closer to the
.05 level, I would be less confident that the results are not due
to random variation. Additionally, a longer time for collection
of data, with a year being recommend, would allow for analysis
of any seasonal trends and also the greater the sample size, the
data would reduce the potential variation because it would
provide for greater accuracy in the results.
262. Hypothesis Testing Paired T
You are in charge of new product development for your
company’s electronic division. One of the key aspects of
electronics is the ability for the battery to extend the time on
the device, specifically battery power. You are considering two
batteries, and old standard battery of a new one that supposedly
has a longer life. You want to test each set of batteries in the
same computer with the same people testing them to minimize
any outside influences in the test. You want to be very
stringent in your analysis so you choose the .01 level of
significance. Is there any difference between the two?
HO:
H1
Decision Rule
Test Statistic:
Decision
Notes:
Participant
New Battery
Old Battery
Bob
45
52
May
41
34
Deno
53
40
Sri
40
263. 38
Pat
43
38
Alexis
43
44
Scott
49
34
Aretha
39
45
Jen
41
28
Ben
43
33
Answer
You are in charge of new product development for your
company’s electronic division. One of the key aspects of
electronics is the ability for the battery to extend the time on
the device, specifically battery power. You are considering two
batteries, and old standard battery of a new one that supposedly
has a longer life. You want to test each set of batteries in the
same computer with the same people testing them to minimize
any outside influences in the test. You want to be very
stringent in your analysis so you choose the .01 level of
significance. Is there any difference between the two?
HO: Battery life of old is equal to battery life of new
H1 Battery life of old is not equal to battery life of new
Decision Rule
Decision Rule: If the probability value calculated is less than
264. .01, we will reject the HO
Test Statistic: .073
Decision: Since the probability value calculated of .073 is
greater than the critical value of .01, we fail to reject HO and
are somewhat confident the battery lives are equal.
Paired T-Test and CI: New Battery, Old Battery
Paired T for New Battery - Old Battery
N Mean StDev SE Mean
New Battery 10 43.70 4.32 1.37
Old Battery 10 38.60 6.98 2.21
Difference 10 5.10 7.94 2.51
95% CI for mean difference: (-0.58, 10.78)
T-Test of mean difference = 0 (vs not = 0): T-Value = 2.03 P-
Value = 0.073
Notes:
Late Fee versus No Late Fee
A blockbuster rental is becoming worried that with the advent
of Netflix etc that customer defection to these new online
rentals may decrease profits. Specifically, customers have
complained about late fees and thus the local managers want to
know if there new policy of no late fees has increased movie
rentals. At the .10 level of significance, what are your
findings?
HO:
266. 10
13
9
Late Fee versus No Late Fee
A blockbuster rental is becoming worried that with the advent
of Netflix etc that customer defection to these new online
rentals may decrease profits. Specifically, customers have
complained about late fees and thus the local managers want to
know if there new policy of no late fees has increased movie
rentals. At the .10 level of significance, what are your
findings?
HO: No Late Fee Movie rentals have decreased or are equal to
Late Fee Rentals
H1No Late Fee Movie rentals have increased over Late Fee
Rentals
Decision Rule If the probability value calculated is less than
.10, we will reject the HO
Test Statistic: .009
Decision: Since the p calculated of .000 is less than .10, we
reject HO and conclude there is strong evidence to conclude
movie rentals have increased.
Paired T-Test and CI: No Late Fee, Late Fee
Paired T for No Late Fee - Late Fee
N Mean StDev SE Mean
No Late Fee 10 12.400 1.430 0.452
Late Fee 10 10.400 2.503 0.792
267. Difference 10 2.000 2.211 0.699
90% lower bound for mean difference: 1.033
T-Test of mean difference = 0 (vs > 0): T-Value = 2.86 P-Value
= 0.009
Notes:
Minitab Instructions
Comparing Two Means with Minitab
I am comparing the mileage of two gasoline’s to see if Chevron
exceeds Shell gasoline in mileage test(DATA IS NOT REAL). I
am interested in is there a difference in the gas mileage between
the two brands.
Step 1
This is not a paired T test rather a standard two sample T test.
Step 2
Since the data is in two columns, click on the Samples in
Different Columns options and then click to enter the two
columns of data for analysis In this case, Chevron was picked
and Shell
Step 3
You have a choice of two graphs and it is personal preference
if you want to use either graph.
Step 4
You choose your alternative hypothesis and also the level of
significance as indicated by the level of confidence. For
example, a .05 level of significance corresponds to a 95%
268. confidence level.
Step 5
Evaluate your graphics
Step 6:
Final Results as seen in Minitab
Two-sample T for Chevron vs Shell
N Mean StDev SE Mean
Chevron 15 29.73 4.92 1.3
Shell 15 25.00 3.95 1.0
Difference = mu (Chevron) - mu (Shell)
Estimate for difference: 4.73333
95% CI for difference: (1.39747, 8.06920)
T-Test of difference = 0 (vs not =): T-Value = 2.91 P-Value =
0.007 DF = 28
Both use Pooled StDev = 4.4599
Arrows indicate how to read print out for the H1 is Chevron is
not equal to Shell
Pay close attention to the sequence listed with the green being
first or like a green light start and red is the second half of the
H1 and red being like a red stop light. (
269. Two Sample T Test Using Summarized Data
We can use a two sample t test to evaluate how well people do
on learning this can be used to evaluate training etc by
administering the same test to two different groups, one with
training and one without training to see if the training will pay
off. In this case, we will simply call them sample one (received
training) and sample two (did not receive training) and then we
gave them the same test
The H 1 choices are less than, not equal to or greater than.
In this example, we are assuming that the first sample mean is
greater than the second sample mean so we choose greater than.
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 20 38.00 3.00 0.67
2 30 35.00 5.00 0.91
Difference = mu (1) - mu (2)
Estimate for difference: 3.00
95% lower bound for difference: 0.91
T-Test of difference = 0 (vs >): T-Value = 2.41 P-Value =
0.010 DF = 48
Both use Pooled StDev = 4.3205
270. HO: Sample one is less than or equal to sample two
H1: sample one is greater than sample two
Decision rule: If P calculated is less than .05, reject HO
Test stat: P value of .010
Decision: Since the p value calculated of .010 less than the
critical value of .05, we reject HO and conclude sample group
one has a higher training score than group two. For if the HO
was true that the training group one has the same or lower score
than training group two, the probability that you would have
randomly found a sample averages of 38 and 35 is is 1.0%,
therefore, it is possible not probable.
Sample Mean
Level of significance/ confidence level
Alternative Hypothesis
Graph Choices
Two columns selected
Select 2 Sample T
271. Enter summarized data
Click on summarized data
Choose H1
_1185526699.txt
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