The document discusses confidence intervals and hypothesis testing. It provides examples of constructing 95% confidence intervals for a population mean using a sample mean and standard deviation. It also demonstrates how to identify the null and alternative hypotheses, determine if a test is right-tailed, left-tailed, or two-tailed, and calculate p-values to conclude whether to reject or fail to reject the null hypothesis based on a significance level of 0.05. Examples include testing claims about population proportions and means.
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Chapter 8: Hypothesis Testing
8.2: Testing a Claim About a Proportion
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Chapter 8: Hypothesis Testing
8.3: Testing a Claim About a Mean
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Chapter 8: Hypothesis Testing
8.2: Testing a Claim About a Proportion
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Chapter 8: Hypothesis Testing
8.3: Testing a Claim About a Mean
Assessing Model Performance - Beginner's GuideMegan Verbakel
Introduction on how to assess the performance of a classifier model. Covers theories (bias-variance trade-off, over/under-fitting), data preparation (train/test split, cross-validation), common performance plots (e.g. ROC curve and confusion matrix), and common metrics (e.g. accuracy, precision, recall, f1-score).
This presentation includes detailed information on Hypothesis testing for large and small samples, for two sample means. Briefed computational procedure with various case studies.
Assessing Model Performance - Beginner's GuideMegan Verbakel
Introduction on how to assess the performance of a classifier model. Covers theories (bias-variance trade-off, over/under-fitting), data preparation (train/test split, cross-validation), common performance plots (e.g. ROC curve and confusion matrix), and common metrics (e.g. accuracy, precision, recall, f1-score).
This presentation includes detailed information on Hypothesis testing for large and small samples, for two sample means. Briefed computational procedure with various case studies.
HW1_STAT206.pdfStatistical Inference II J. Lee Assignment.docxwilcockiris
HW1_STAT206.pdf
Statistical Inference II: J. Lee Assignment 1
Problem 1. Suppose the day after the Drexel-Northeastern basketball game, a poll of 1000 Drexel students
was conducted and it was determined that 850 out of the 1000 watched the game (live or on television).
Assume that this was a simple random sample and that the Drexel undergraduate population is 20000.
(a) Generate an unbiased estimate of the true proportion of Drexel undergraduate students who watched
the game.
(b) What is your estimated standard error for the proportion estimate in (a)?
(c) Give a 95% confidence interval for the true proportion of Drexel undergraduate students who watched
the game.
Problem 2. (Exercise 18 in Chapter 7 of Rice) From independent surveys of two populations, 90% con-
fidence intervals for the population means are conducted. What is the probability that neither interval
contains the respective population mean? That both do?
Problem 3. (Exercise 23 in Chapter 7 of Rice)
(a) Show that the standard error of an estimated proportion is largest when p = 1/2.
(b) Use this result and Corollary B of Section 7.3.2 (also, on Page 17 of the lecture notes) to conclude that
the quantity
1
2
√
N − n
N(n − 1)
is a conservative estimate of the standard error of p̂ no matter what the value of p may be.
(c) Use the central limit theorem to conclude that the interval
p̂ ±
√
N − n
N(n − 1)
contains p with probability at least .95.
HW2_STAT206.pdf
Statistical Inference II: J. Lee Assignment 2
Problem 1. The following data set represents the number of NBA games in January 2016, watched by 10
randomly selected student in STAT 206.
7, 0, 4, 2, 2, 1, 0, 1, 2, 3
(a) What is the sample mean?
(b) Calculate sample variance.
(c) Estimate the mean number of NBA games watched by a student in January 2016.
(d) Estimate the standard error of the estimated mean.
Problem 2. True or false? Tell me why for the false statements.
(a) The center of a 95% confidence interval for the population mean is a random variable.
(b) A 95% confidence interval for µ contains the sample mean with probability .95.
(c) A 95% confidence interval contains 95% of the population.
(d) Out of one hundred 95% confidence intervals for µ, 95 will contain µ.
Problem 3. An investigator quantifies her uncertainty about the estimate of a population mean by reporting
X ± sX . What size confidence interval is?
Problem 4. For a random sample of size n from a population of size N, consider the following as an
estimate of µ:
Xc =
n∑
i=1
ciXi,
where the ci are fixed numbers and X1, . . . ,Xn are the sample. Find a condition on the ci such that the
estimate is unbiased.
Problem 5. A sample of size 100 has the sample mean X = 10. Suppose the we know that the population
standard deviation σ = 5. Find a 95% confidence interval for the population mean µ.
Problem 6. Suppose the we know that the population standard deviation σ = 5. Then how large should a
sample be to estimate the popula.
1. Illustrate point and interval estimations.
2. Distinguish between point and interval estimation.
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Chapter 7: Estimating Parameters and Determining Sample Sizes
7.1: Estimating a Population Proportion
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Chapter 7: Estimating Parameters and Determining Sample Sizes
7.1: Estimating a Population Proportion
Statistics practice for finalBe sure to review the following.docxdessiechisomjj4
Statistics practice for final
Be sure to review the following and have this information handy when taking Final GHA:
· Calculating z alpha/2 and t alpha/2 on Tables II and IV
· Find sample size for estimating population mean. Formula 8.3 p. 321 OCR.
· Stating H0 and H1 claims about variation and about the mean. Chapter 9 OCR.
· Type I and Type II errors p. 345 OCR.
· Confidence Interval for difference between two population means. Chapter 10 OCR p. 428
· Pooled sample standard deviation. Chapter 10 OCR p. 397
· What do Chi-Square tests measure? How are their degrees of freedom calculated? Chapter 12 OCR
· Find F test statistic using One-Way ANOVA.xls Be sure to enable editing and change values to match your problem. One-Way ANOVA.xls
· Find equation of regression line used to predict. To do on Excel, go to a blank worksheet, enter x values in one column and their matching y values in another column. Click Insert – Select Scatterplot. Right click any one of the points (diamonds) on the graph. Left click “Add a Trendline.” Check “Display Equation on Chart” box. Regression equation will appear on chart. Try it here with No. 20 below.
Practice Problems
Chapter 8 Final Review
1) In which of the following situations is it reasonable to use the z-interval
procedure to obtain a confidence interval for the population mean?
Assume that the population standard deviation is known.
A. n = 10, the data contain no outliers, the variable under consideration is
not normally distributed.
B. n = 10, the variable under consideration is normally distributed.
C. n = 18, the data contain no outliers, the variable under consideration is
far from being normally distributed.
D. n = 18, the data contain outliers, the variable under consideration is
normally distributed.
Find the necessary sample size.
2) The weekly earnings of students in one age group are normally
distributed with a standard deviation of 10 dollars. A researcher wishes to
estimate the mean weekly earnings of students in this age group. Find the
sample size needed to assure with 95 percent confidence that the sample
mean will not differ from the population mean by more than 2 dollars.
Find the specified t-value.
3) For a t-curve with df = 6, find the two t-values that divide the area under
the curve into a middle 0.99 area and two outside areas of 0.005.
Provide an appropriate response.
4) Under what conditions would you choose to use the t-interval procedure
instead of the z-interval procedure in order to obtain a confidence
interval for a population mean? What conditions must be satisfied in
order to use the t-interval procedure?
CHAPTER 8 Answers
1) B
2) 97
3) -3.707, 3.707
4) When the population standard deviation is unknown, the t-interval procedure is used instead of the
z-interval procedure. The t-interval procedure works provided that the population is normally
distributed or the.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Ethnobotany and Ethnopharmacology:
Ethnobotany in herbal drug evaluation,
Impact of Ethnobotany in traditional medicine,
New development in herbals,
Bio-prospecting tools for drug discovery,
Role of Ethnopharmacology in drug evaluation,
Reverse Pharmacology.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
2. Twenty five samples from the same population gave these 95% confidence intervals, 95% of all samples give an interval that contains the population mean u.
3. In 95% of all samples, x-bar lies within ± 9 of the unknown population mean µ. Also µ lies within ± 9 of x-bar of the samples.
4. To say that x-bar is a 95% confidence interval for the population mean µ is to say that in repeated samples, 95% of these intervals capture the µ
5. The central probability 0.8 under a standard normal curve lies between -1.28 and + 1.28. That is, there is area .1 to the left and to the right of the curve.
6. The central probability C under a standard normal curve lies between –z* and + z* has area (1-C)/2 to its right under the curve, we call it the upper (1-C)/2 critical value.
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11. Confidence Level C Single Tail Area (1-C)/2 Critical z value 90% .05 1.645 95% .025 1.960 99% .005 2.576
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13. Confidence Interval for p Single Proportion Confidence Interval for µ Single Mean ( σ known) Conditions Conditions If an SRS of size n is chosen from a population with unknown proportion p, then a level C Confidence Interval for p is: If an SRS of size n is chosen from a population with unknown proportion, µ then a level C Confidence Interval for µ is: p-hat ± z critical value ( ((p(1-p)/ n) X-bar ± z critical value ( σ /
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22. Tests of Significance: We have learned that Confidence Intervals can be used to estimate a parameter. Often in statistics we want to use sample data to determine whether or not a claim or hypothesis about a parameter is plausible. A test of significance is a procedure in which we can use sample data to test such a claim. We will focus on tests about a population mean μ and proportions p.
28. c) The standard deviation of IQ scores of actors is equal to 15. Practice : Identify the Null and Alternative Hypothesis. Express the corresponding null and alternative hypotheses in symbolic form.
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36. To determine the significance at a set level, alpha, one needs to calculate the P-value for the observed statistic. Careful consideration of the test statistic, z, can also be used to determine significance:
37. P -Value The P -value (or p -value or probability value ) is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true. The null hypothesis is rejected if the P -value is very small, such as 0.05 or less.
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43. Critical Region The critical region (or rejection region ) is the set of all values of the test statistic that cause us to reject the null hypothesis.
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45. Practice: Finding P -values. First determine whether the given conditions result in a right-tailed test, a left-tailed test, or a two-tailed test, then find the P -values and state a conclusion about the null hypothesis. a) With a claim of p > 0.25, the test is right-tailed. Because the test is right-tailed, the graph shows that the P -value is the area to the right of the test statistic z = 1.18. Using Table A and find that the area to the right of z = 1.18 is 0.1190. The P -value is 0.1190 is greater than the significance level = 0.05, so we fail to reject the null hypothesis.
46. Practice: Finding P -values. First determine whether the given conditions result in a right-tailed test, a left-tailed test, or a two-tailed test, then find the P -values and state a conclusion about the null hypothesis. b) A significance level of = 0.05 is used in testing the claim that p 0.25 , and the sample data result in a test statistic of z = 2.34 .
47. Practice: Finding P -values. First determine whether the given conditions result in a right-tailed test, a left-tailed test, or a two-tailed test, then find the P -values and state a conclusion about the null hypothesis. b) With a claim of p 0.25, the test is two-tailed. Because the test is two-tailed, and because the test statistic of z = 2.34 is to the right of the center, the P-value is twice the area to the right of z = 2.34. Using Table A and find that the area to the right of z = 2.34 is 0.0096, so P-value = 2 x 0.0096 = 0.0192. The P-value of 0.0192 is less than or equal to the significance level, so we reject the null hypothesis.
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51. Example 4: We are given a data set of 106 body temperatures having a mean of 98.20°F. Assume that the sample is a simple random sample and that the population standard deviation is known to be 0.62°F. Use a 0.05 significance level to test the common belief that the mean body temperature of healthy adults is equal to 98.6°F. Use the P -value method. This is a two-tailed test and the test statistic is to the left of the center, so the P -value is twice the area to the left of z = –6.64. We refer to Table A to find the area to the left of z = –6.64 is 0.0001, so the P -value is 2(0.0001) = 0.0002. = z = x – µ x n 98.2 – 98.6 = –6.64 106 H 0 : = 98.6 H 1 : 98.6 = 0.05 x = 98.2 = 0.62
52. Example 4: We are given a data set of 106 body temperatures having a mean of 98.20°F. Assume that the sample is a simple random sample and that the population standard deviation is known to be 0.62°F. Use a 0.05 significance level to test the common belief that the mean body temperature of healthy adults is equal to 98.6°F. Use the P -value method. z = –6.64 H 0 : = 98.6 H 1 : 98.6 = 0.05 x = 98.2 = 0.62
53. Example 4: We are given a data set of 106 body temperatures having a mean of 98.20°F. Assume that the sample is a simple random sample and that the population standard deviation is known to be 0.62°F. Use a 0.05 significance level to test the common belief that the mean body temperature of healthy adults is equal to 98.6°F. Use the P -value method. z = –6.64 Because the P -value of 0.0002 is less than the significance level of = 0.05, we reject the null hypothesis. There is sufficient evidence to conclude that the mean body temperature of healthy adults differs from 98.6°F. H 0 : = 98.6 H 1 : 98.6 = 0.05 x = 98.2 = 0.62
54. Example 5: A survey of n = 880 randomly selected adult drivers showed that 56%(or p-hat = 0.56) of those respondents admitted to running red lights. Find the value of the test statistic for the claim that the majority of all adult drivers admit to running red lights.
55. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. np = (880)(0.5) = 440 5 n(1-p) = (880)(0.5) = 440 5
56. Example 5: In an article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the P -value Method. H 0 : p = 0.5 H 1 : p > 0.5 = 0.05 Referring to Table A, we see that for values of z = 3.50 and higher, we use 0.9999 for the cumulative area to the left of the test statistic. The P-value is 1 – 0.9999 = 0.0001. p(1-p) n p – p z = 0.56 – 0.5 (0.5)(0.5) 880 = = 3.56
57. Interpretation: In an article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the P -value Method. H 0 : p = 0.5 H 1 : p > 0.5 = 0.05 We know from previous chapters that a z score of 3.56 is exceptionally large. The corresponding P -value of 0.0001 is less than the significance level of = 0.05, we reject the null hypothesis. There is sufficient evidence to support the claim. p(1-p) n p – p z = 0.56 – 0.5 (0.5)(0.5) 880 = = 3.56
58. Interpretation : In an article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the P -value Method. H 0 : p = 0.5 H 1 : p > 0.5 = 0.05 z = 3.56
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61. Traditional method : Reject H 0 if the test statistic falls within the critical region. Fail to reject H 0 if the test statistic does not fall within the critical region. Decision Criterion
62. P -value method : Reject H 0 if P -value (where is the significance level, such as 0.05). Fail to reject H 0 if P -value > . ** this is the method you will use Decision Criterion**
63. Decision Criterion Confidence Intervals : Because a confidence interval estimate of a population parameter contains the likely values of that parameter, reject a claim that the population parameter has a value that is not included in the confidence interval.
page 374 of text The term ‘accept’ is somewhat misleading, implying incorrectly that the null has been proven. The phrase ‘fail to reject’ represents the result more correctly.