inferential statistics, statistical inference, language technology, interval estimation, confidence interval, standard error, confidence level, z critical value, confidence interval for proportion, confidence interval for the mean, multiplier,
Lecture 6 Point and Interval Estimation.pptxshakirRahman10
Point and Interval Estimation:
Objectives:
Apply the basics of inferential statistics in terms of point estimation.
Compute point and estimation of population means and confidence interval.
Interpret the results of point and interval estimation.
Estimation:
Estimating the value of parameter from the sample:
An aspect of inferential statistics.
Why to estimate: Population is large enough so we
can only estimate.
Types of estimation:
Point Estimation:
A specified number value (single value) that is an estimate of a population parameter. The point estimate of the population mean µ is the sample mean.
Interval Estimate:
Range of values to estimate about population parameter.
Confidence Interval Estimation:
Range of values to estimate about population parameter.
May contain the parameter or not (Degree of confidence).
Ranges between two values.
Example:
Age (in years) 4 BScN students: 20<µ < 25 or (22.5 +2.5)
FORMULA:
Point estimate (x) + Critical Value x Standard Error.
Confidence Interval is a particular interval of estimate.
Given that sample size is large, the 95% of the sample means taken from same population and same sample size will fall in + 1.96 SD of the population mean.
Three commonly used Confidence Intervals are 90%, 95% (by default) , and 99%.
Why not too small or too large confidence intervals?
Too wide: 99.9% Interval too broad
Too narrow: 80 % More uncertainty to have population mean.
The 99% of the sample means taken from same population and
same sample size will fall in + 2.575 SD of the population mean.
Interpretation:
99% probability that interval will enclose population parameter and 1% chance that it will not have population parameter.
Level of confidence: The level of certainty that the interval will have the true population mean.
Chances of Error: Chances that the interval will not cater the true parameter.
Sum of level of confidence and chances of error =100%
Multiple Linear Regression II and ANOVA IJames Neill
Explains advanced use of multiple linear regression, including residuals, interactions and analysis of change, then introduces the principles of ANOVA starting with explanation of t-tests.
inferential statistics, statistical inference, language technology, interval estimation, confidence interval, standard error, confidence level, z critical value, confidence interval for proportion, confidence interval for the mean, multiplier,
Lecture 6 Point and Interval Estimation.pptxshakirRahman10
Point and Interval Estimation:
Objectives:
Apply the basics of inferential statistics in terms of point estimation.
Compute point and estimation of population means and confidence interval.
Interpret the results of point and interval estimation.
Estimation:
Estimating the value of parameter from the sample:
An aspect of inferential statistics.
Why to estimate: Population is large enough so we
can only estimate.
Types of estimation:
Point Estimation:
A specified number value (single value) that is an estimate of a population parameter. The point estimate of the population mean µ is the sample mean.
Interval Estimate:
Range of values to estimate about population parameter.
Confidence Interval Estimation:
Range of values to estimate about population parameter.
May contain the parameter or not (Degree of confidence).
Ranges between two values.
Example:
Age (in years) 4 BScN students: 20<µ < 25 or (22.5 +2.5)
FORMULA:
Point estimate (x) + Critical Value x Standard Error.
Confidence Interval is a particular interval of estimate.
Given that sample size is large, the 95% of the sample means taken from same population and same sample size will fall in + 1.96 SD of the population mean.
Three commonly used Confidence Intervals are 90%, 95% (by default) , and 99%.
Why not too small or too large confidence intervals?
Too wide: 99.9% Interval too broad
Too narrow: 80 % More uncertainty to have population mean.
The 99% of the sample means taken from same population and
same sample size will fall in + 2.575 SD of the population mean.
Interpretation:
99% probability that interval will enclose population parameter and 1% chance that it will not have population parameter.
Level of confidence: The level of certainty that the interval will have the true population mean.
Chances of Error: Chances that the interval will not cater the true parameter.
Sum of level of confidence and chances of error =100%
Multiple Linear Regression II and ANOVA IJames Neill
Explains advanced use of multiple linear regression, including residuals, interactions and analysis of change, then introduces the principles of ANOVA starting with explanation of t-tests.
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.
JKN 10 Inference 2 Populations.Consider the following set of d.docxchristiandean12115
JKN 10 Inference 2 Populations.
Consider the following set of data.
Pairs
1
2
3
4
5
Sample A
9
4
3
5
3
Sample B
3
8
2
7
1
(a) Find the paired differences, d = A - B, for this set of data.
(d1)
(d2)
(d3)
(d4)
(d5)
(b) Find the mean d of the paired differences. (Give your answer correct to one decimal place.)
(c) Find the standard deviation sd of the paired differences. (Give your answer correct to two decimal places.)
Salt-free diets are often prescribed to people with high blood pressure. The following data values were obtained from an experiment designed to estimate the reduction in diastolic blood pressure as a result of consuming a salt-free diet for 2 weeks. Eight subjects had their blood pressure measured and then ate a salt free diet for two weeks and had their blood pressure measured again. Assume diastolic readings to be normally distributed.
Before
99
105
93
102
100
108
107
97
After
92
102
91
94
96
98
100
93
(a) The proper TI-83 program to use to compute the confidence interval for the mean reduction in blood pressure is:
(b) Find the 98% confidence interval for the mean reduction. (Give your answers correct to two decimal places.)
Lower Limit
Upper Limit
(c) Which of the following statements is true about the confidence interval? (More than one may apply)
We are 98% confident that the true mean of the individual differences in blood pressure is in the intervalIf we took 100 samples and constructed 100 confidence intervals for the mean of the individual differences, approximately 98 of them would contain the true mean of the individual differencesWe are 98 % confident that the mean of the after data minus the mean of the before data is in the intervalWe are 98 % confident that the mean of the before data minus the mean of the after data is in the interval
An experiment was designed to estimate the mean difference in weight gain for pigs fed ration A as compared with those fed ration B. Eight pairs of pigs were used. The pigs within each pair were litter-mates. The rations were assigned at random to the two animals within each pair. The gains (in pounds) after 45 days are shown in the following table. Assuming weight gain is normal, find the 99% confidence interval estimate for the mean of the differences μd, where d = ration A - ration B. (Give your answers correct to two decimal places.)
Litter
1
2
3
4
5
6
7
8
Ration A
56
40
60
59
43
40
50
46
Ration B
54
30
50
56
37
36
42
40
Lower Limit
Upper Limit
State the null hypothesis, Ho, and the alternative hypothesis, Ha, that would be used to test these claims.
(a) There is an increase in the mean difference between post-test and pretest scores.(d=post-test scores - pretest scores)
Ho: μd
0
Ha: μd
0
(b) Following a special training session, it is believed that the mean of the difference in performance scores will not be zero.
Ho: μd
0
Ha: μd
0
(c) On average, there is no difference between the readings from two inspectors on .
ELEMENTS OF STATISTICS / TUTORIALOUTLET DOT COMalbert0076
Unit 3 Problem Set NAME: Elements of Statistics--FHSU Virtual College--Spring 2017
REMEMBER, these are assessed preparatory problems related to the content of Unit 3. The Unit 3 Exam will consist of similar types of
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Elementary Statistics Practice Test 1
Module 1: Chapters 1-3
Chapter 1: Introduction to Statistics.
Chapter 2: Exploring Data with Tables and Graphs.
Chapter 3: Describing, Exploring, and Comparing Data.
Part 1 of 16 -Question 1 of 231.0 PointsThe data presented i.docxodiliagilby
Part 1 of 16 -
Question 1 of 23
1.0 Points
The data presented in the table below resulted from an experiment in which seeds of 5 different types were planted and the number of seeds that germinated within 5 weeks after planting was recorded for each seed type. At the .01 level of significance, is the proportion of seeds that germinate dependent on the seed type?
Seed Type
Observed Frequencies
Germinated
Failed to Germinate
1
31
7
2
57
33
3
87
60
4
52
44
5
10
19
Reset Selection
Question 2 of 23
1.0 Points
A company operates four machines during three shifts each day. From production records, the data in the table below were collected. At the .05 level of significance test to determine if the number of breakdowns is independent of the shift.
Machine
Shift
A
B
C
D
1
41
20
12
16
2
31
11
9
14
3
15
17
16
10
Reset Selection
Part 2 of 16 -
Question 3 of 23
1.0 Points
In choosing the “best-fitting” line through a set of points in linear regression, we choose the one with the:
Reset Selection
Question 4 of 23
1.0 Points
A single variable X can explain a large percentage of the variation in some other variable Y when the two variables are:
Reset Selection
Question 5 of 23
1.0 Points
A correlation value of zero indicates.
Reset Selection
Part 3 of 16 -
Question 6 of 23
1.0 Points
An investor wants to compare the risks associated with two different stocks. One way to measure the risk of a given stock is to measure the variation in the stock’s daily price changes.
In an effort to test the claim that the variance in the daily stock price changes for stock 1 is different from the variance in the daily stock price changes for stock 2, the investor obtains a random sample of 21 daily price changes for stock 1 and 21 daily price changes for stock 2.
The summary statistics associated with these samples are: n
1
= 21, s
1
= .725, n
2
= 21, s
2
= .529.
If you compute the test value by placing the larger variance in the numerator, at the .05 level of significance, would you conclude that the risks associated with these two stocks are different?
Reset Selection
Question 7 of 23
1.0 Points
Two independent samples of sizes n
1
= 50 and n
2
= 50 are randomly selected from two populations to test the difference between the population means,
. The sampling distribution of the sample mean difference,
is:
Reset Selection
Part 4 of 16 -
Question 8 of 23
1.0 Points
Suppose that the mean time for a certain car to go from 0 to 60 miles per hour was 7.7 seconds. Suppose that you want to test the claim that the average time to accelerate from 0 to 60 miles per hour is longer than 7.7 seconds. What would you use for the alternative hypothesis?
Reset Selection
Question 9 of 23
1.0 Points
A two-tailed test is one where:
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Question 10 of 23
1.0 Points
Which of the following values is not typically used for
?
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Part 5 of 16 -
Question 11 of 23
1.0 Points
From a sample of 500 items, 30 wer.
Categorical Data and Statistical AnalysisMichael770443
In this presentation, we will introduce two tests and hypothesis testing based on it, and different non-parametric methods such as the Kolmogorov-Smirnov test, the Wilcoxon’s signed-rank test, the Mann-Whitney U test, and the Kruskal-Wallis test.
Researchers use several tools and procedures for analyzing quantitative data obtained from different types of experimental designs. Different designs call for different methods of analysis. This presentation focuses on:
T-test
Analysis of variance (F-test), and
Chi-square test
The Art Pastor's Guide to Sabbath | Steve ThomasonSteve Thomason
What is the purpose of the Sabbath Law in the Torah. It is interesting to compare how the context of the law shifts from Exodus to Deuteronomy. Who gets to rest, and why?
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
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.
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.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
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.
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.
2. Purpose of the Chi Squared Test The purpose of the chi squared test is to see whether observed experimental data is a ‘good fit’ with theoretical expected results. The χ 2 statistic is calculated in the following way:
4. Notes on the Chi Squared Distribution 1.) ν (the number of degrees of freedom) is calculated from the number of classes – the number of restrictions. 2.) A restriction is defined as any value that is derived from the observed data set. 3.) The chi squared distribution is continuous and thus offers poor approximation when dealing with small frequencies. When calculating χ 2 we have to combine any classes which contain expected frequencies of less than 5 elements. 4.) As with most statistical distributions, we do not need to concern ourselves with calculating by hand as all critical values are tabulated for easy reference.
5. Lesson 1 - Example Question I The table below shows the results when a die is rolled 120 times. Conduct a chi squared test to see whether the die is fair or not at the 5% significance level. 24 20 18 14 29 15 Freq 6 5 4 3 2 1 Score
6. Lesson 1 - Example Question II The table below shows the results of an experiment in which four coins thrown 160 times and the number of heads recorded. a.) Fit a Bin (4, ½) distribution to the data. b.) Test the goodness of fit of the Bin (4, ½) model using a chi squared test at the 5% significance level. 10 35 54 46 15 Freq 4 3 2 1 0 Score
7. Practice Questions Statistics 3 and 4 by Jane Miller Page 121, Exercise 5A Questions 1 and 3
8.
9.
10. Lesson 3 - Example Question The table below shows the number of calls arriving at a switchboard in time intervals of 5 minutes. Test at the 5% significance level whether the Poisson distribution provides a good model for this data. 0 2 4 23 71 Freq 4 or more 3 2 1 0 No
11. Practice Questions Statistics 3 and 4 by Jane Miller Page 121, Exercise 5A Questions 11 (part a only) (Poisson) 2 (Ratio) 12 (Geometric) 13 (Poisson)
12.
13. Lesson 4 - Example Question The height in centimetres gained by a conifer in its first year after planting is denoted by the random variable H. The value of H is measured for a random sample of 86 conifers and the results obtained are summarised in the table below. Assuming that H is modelled by a N(50, 15 2 ) distribution, test at the 5% level, the goodness of fit of the model. 12 18 28 18 10 Obs Freq >65 55-65 45-55 35-45 <35 H
14. Practice Questions Statistics 3 and 4 by Jane Miller Page 128, Exercise 5B Question 1 onwards
15.
16. Degrees of Freedom The number of degrees of freedom in a h x k contingency table is given by ν = (h – 1) x (k – 1).
17. Lesson 5 - Example Question I Is income level independent of method of transport? 693 129 102 462 Total 266 29 32 205 Large 312 64 49 199 Average 115 36 21 58 Small Income Level Total Self Public Car Method of Transport
18. Lesson 5 - Example Question II A university sociology department believes that students with a good grade in A Level General Studies tend to do well on sociology degree courses. To check this it has collected information on a random sample of 100 who had just graduated and who also had taken general studies at A Level. The students performance in General Studies was divided into two categories, those with grades A or B and ‘others’. Their degrees were recorded as Class I, II, III or fail. The data is given in the table below. Test at the 1% level, the hypothesis that degree performance is independent of A level performance in General Studies. 100 5 30 50 15 Total 60 4 24 28 4 Others 40 1 6 22 11 Grade A or B Total Fail Class III Class II Class I
19. Yates Correction χ 2 is a continuous distribution whilst χ 2 calc is not. In the case of a 2 x 2 contingency table for which ν = 1, the agreement between the two distributions can be improved by applying a continuity correction called Yates Correction. This involves reducing each value of |O – E| by 0.5
20. Lesson 5 - Example Question III A random sample of 930 companies quoted on the stock exchange revealed the information summarised in the table below, which shows the distribution of these companies classified according to two attributes. In this table, D indicates that the company has diversified its product range during the previous financial year, and P indicates that there has been a significant rise in profits during the previous financial year. The null hypothesis is that D and P are independent. Show that this can rejected at all reasonable significance levels. 377 299 Not P 106 148 P Not D D