CH-9-10 Estimation and Hypothesis Testing I: Single Sample T-Test and Z-Test – Introduction to Statistics for Social Sciences : Book Guide

This Document Contains Chapters 9 to 10 Chapter 9 Estimation and Hypothesis Testing I: Single Sample T-Test and Z-Test Learning Objectives: 1. Describe the importance of a single sample test of a mean and proportion. 2. Describe when to use a z-test versus a t-test when testing a hypothesis about a single mean. 3. Prepare a hypothesis for testing using a single sample test. 4. Conduct a single sample z-test of a mean. 5. Conduct a single sample t-test of a mean. 6. Conduct a single sample z-test of a proportion. 7. Describe why you might favour a two-tailed test over a one-tailed test. 8. Report the p-values for a z-test and/or a t-test. Chapter Summary In this chapter, students transition their learning to statistical hypothesis testing. Specifically this chapter deals with hypothesis testing with a single sample mean and single sample proportion. Key Formulas The following represent the key formulas for this chapter. PowerPoint slides are provided for each chapter. In addition to these slides, a PDF file containing only the formulas are also provided. Single sample z-test for a mean Standard error of the mean Single sample t-test for a mean Estimated standard error of the mean Single sample z-test for a proportion This Document Contains Chapters 9 to 10 Estimated standard error of the proportion Interactive Figures: The textbook contains interactive figures. You may wish to use these in a lecture. Students also have access to these. For this chapter, there is one interactive figure. 1. Interactive calculator for the z-test and t-test for the mean and z-test for the proportion. These interactive figures can be found on in the eBook and the Library under Chapter 9 Resources. Typical Lecture Material We have provided two sample lectures below. You may wish to add in additional discipline specific information to make these more relevant to your students. Lecture 1: Objective: To understand when to use a z-distribution versus a t-distribution to estimate critical values and complete hypothesis testing. Review the following concept table with your students; and help them to fill in the definitions of each statistical concept. The definitions that the students should come up with are in italics. Statistical Concept Defintion Hypothesis testing with a single sample mean The statistical evaluation of a hypothesis whereby we compare a single sample mean to a known mean which is often, but not always, the population mean. Hypothesis testing with a single sample proportion The statistical evaluation of a hypothesis whereby we compare a single sample proportion to a known proportion which is often, but not always a population proportion. A single sample z-test of a mean. The comparison of a single sample mean to a known population mean when the population standard deviation is known. A single sample t-test of a mean. The comparison of a single sample mean to a known population mean when the population standard deviation is unknown. A single sample z-test of a proportion. The comparison of a single sample proportion to a known or desired population proportion. Example 1: This exercise helps students to remember when a single sample z-test or single sample t-test is used. Ask your students to fill in the following table components based on the supplementary information provided. (Answers are in italics) Sample Value of Interest Population Standard Deviation (𝜎) Statistical Test Sample Mean (𝑥̅) Known Single- sample z-test for a mean Sample Mean (𝑥̅) Unknown Single-sample t-test for a mean. Sample Proportion (𝑝̂) Unknown Single-sample z-test for a proportion. Example 2: Single sample t-test when the population standard deviation is known. Read the following scenario to your students: Access to child and youth mental health services has become a recognized priority within the Ministry of Health. Currently, the mean waiting time for non-crisis services is 60 days from the time of referral; with a standard deviation of 10 days. Although you have implementing an advanced triaging system, you are unsure if this has actually made a difference. Your supervisor has asked you to empirically test whether the triaging system has changed the mean wait time of 60 days. You hypothesize that the mean wait time is different (two-tailed hypothesis); and you randomly select 150 non-crisis cases. You want a confidence level of 99%, so your alpha value (α) is 0.01. After collecting the data, you find that the mean wait time is 50 days. Ask your students to complete the hypothesis steps. The answers follow each step below. 1. Define the null and alternative hypothesis using words and statistical notation. H0 : μ = 60 days HA : μ ≠ 60 days 2. Define the sampling distribution and critical values The mean of the sampling distribution (𝜇𝑥̅) is equal to the population mean (μ) of 60 days. The standard error of the mean is: 𝜎𝑥̅= 𝜎 √𝑛 = 10 √150 = 10 12.25 = 0.82 The sampling distribution would look like the following: Critical value of alpha at .01 (two-sided) = ± 2.576 3. Calculate the test-statistic using the sample data 𝑧 = 𝑥̅− 𝜇0 ( 𝜎 √𝑛) = 50 − 60 ( 10 √150) = −10 0.82 = −12.20 4. Make the decision regarding the hypothesis Our z-statistic is -12.20, meaning our sample mean is 12.20 standard errors less than the population mean of 60 days. The two-tailed critical value of .01 (meaning 1% chance of rejecting a false null) is + 2.576. Given our z-statistic is past our critical value (-12.20 < -2.576), we reject the null hypothesis that μ = 60. 5. Interpret the results Based on our sample, there appears to be a statistically significant decrease in wait times. The sample mean of 50 days is significantly lower than the population mean of 60 days at an alpha (α) level of .01. Lecture 2: Example 1: Single sample t-test when the population standard deviation is unknown. Read the following scenario to your students: More often than not, we do not know the population standard deviation that we wish to compare our own sample to. Using the same scenario from the previous lecture, assume that we do not know the population standard deviation, but rather we do know the sample standard deviation, which is 12 days. To recap, we have a sample mean of 50 days, a sample standard deviation of 12 days and a sample size of 150. You want a confidence level of 99%, so your alpha value (α) is 0.01. Ask your students to complete the hypothesis steps. The answers follow each step below. 1. Define the null and alternative hypothesis using words and statistical notation. H0 : μ = 60 days HA : μ ≠ 60 days 2. Define the sampling distribution and critical values The mean of the sampling distribution (𝜇𝑥̅) is equal to the population mean (μ) of 60 days. The standard error of the mean is: 𝑠𝑥̅= 𝑠 √𝑛 = 12 √150 = 12 12.25 = 0.98 The sampling distribution would look like the following: Critical value of alpha at .01 (two-sided) = ± 2.576 3. Calculate the test-statistic using the sample data 𝑡 = 𝑥̅− 𝜇0 ( 𝑠 √𝑛 ) = 50 − 60 ( 12 √150 ) = −10 . 98 = −10.20 4. Make the decision regarding the hypothesis Our z-statistic is -10.20, meaning our sample mean is 10.20 standard errors less than the population mean of 60 days. The two-tailed critical value of .01 (meaning 1% chance of rejecting a false null) is + 2.576. Given our t-statistic is past our critical value (-10.20 < -2.576), we reject the null hypothesis that μ = 60. 5. Interpret the results Based on our sample, there appears to be a statistically significant decrease in wait times. The sample mean of 50 days is significantly lower than the population mean of 60 days at an alpha (α) level of .01. Example 2: Read the following to your students: Last year, 49% of the phone calls to your Youth Help Line were answered within three telephone rings. After adding some additional volunteers, you wish to test if there is a significant difference in the proportion of calls answered within 3 rings. You hypothesize that there is a difference. After collecting the data from 130 phone calls, you find that the proportion of calls answered within 3 rings has increased to 54%. You now wish to test your hypothesis using an alpha value of 0.05. Ask your students to complete the hypothesis steps. The answers follow each step below. 1. Define the null and alternative hypothesis using words and statistical notation. H0 : p = 0.49 HA : p ≠ 0.49 2. Define the sampling distribution and critical values The standard error of the mean is: 𝜎𝑝 ̂ = √ 𝑝0(1 − 𝑝0) 𝑛 = √ . 49(1 − .49) 130 = √ . 25 130 = 0.04 The sampling distribution would look like the following: Critical value of alpha at .05 (two-sided) = ± 1.96 3. Calculate the test-statistic using the sample data 𝑧 = 𝑝̂− 𝑝0 √𝑝0(1 − 𝑝0) 𝑛 = . 54 − .49 √. 49(1 − .49) 130 = 0.05 0.04 = 1.25 4. Make the decision regarding the hypothesis Our z-statistic is 1.25, meaning our sample mean is 1.25 standard errors higher than the population proportion of .49. The two-tailed critical value of .05 is + 1.96. Given our z-statistic does not exceed the critical value (1.25 570 This is a t-test, as population standard deviation is not known. Test statistic = t = 3.58 t critical value is 1.729 Test statistic is greater than the critical value. Hence reject H0. There is sufficient evidence to believe that the mean GMAT of the coaching centre is greater than the mean GMAT score. Exercise 9-2 A simple random sample of size 25 is drawn from a population. The sample mean is found to be 40. The population standard deviation is 5. Test the hypothesis that the mean is different from 45. Feedback: H0 : µ = 45 H1 : µ ≠ 45 This is a z-test, as population standard deviation is known. Test statistic = Z = 1 P-value = 2* P(Z>1) = 2*0.1587 = 0.3174 P-value is greater than 0.05. Hence fail to reject H0. There is not enough evidence to believe that the mean is different from 45. Exercise 9-3 A sociologist wishes to conduct a poll to estimate the percentage of Canadians who prefer gun control. He took a random sample of 200 people and noted that 54 favoured privatisation of health care. Test the hypothesis, at 0.05 level of significance, that the proportion of those who prefer gun control of health care is different from the assumed 30%. Feedback: H0 : p = 0.30 H1 : p ≠ 0.30 This is a z-test, as it is a test for proportion. Test statistic = Z = 0.93 Z critical value = 1.96. As Test statistic is less than the critical Solutions to SPSS Exercises Question 9-1 Clay234 is a new type of cold remedy. The following data represent the number of hours it takes a patient to have noticeable relief from their cold symptoms. 10 20 7 19 32 6 23 16 20 35 5 5 11 15 33 12 31 12 22 11 15 a) Recent studies would indicate that the fastest remedies available are on average 13 hours to relief. Does the data indicate that the average is different from 13? Use alpha = .01 Ans: One-Sample Test t df Sig. (2-tailed) Mean Difference 95% Confidence Interval of the Difference Lower Upper VAR00005 2.017 20 .057 4.14286 -.1425 8.4282 Since the significance is greater than alpha then the null is maintained and there is no difference in the cold remedies. LO: 5 Page: 251 Equations from Chapter 9 Scott R. Colwell and Edward M. Carter c 2012 Equation 9.1: Single Sample z-Test for a Mean z = x¯ − µ0 σx¯ Where: z = z-statistic x¯ = sample mean µ0 = the population mean σx¯ = standard error of the mean Equation 9.2: Standard Error of the Mean σx¯ = √σ n Where: σx¯ = standard error of the mean σ = population standard deviation n = sample size Equation 9.3: Single Sample z-Test for a Mean z = x¯ − µ0 √σ n Where: z = z-statistic x¯ = sample mean µ0 = the population mean σ = population standard deviation n = sample size Equation 9.6: Single Sample t-Test for a Mean t = x¯ − µ0 sx¯ Where: t = t-statistic x¯ = sample mean µ0 = the population mean sx¯ = estimated standard error of the mean Equation 9.7: Estimated Standard Error of the Mean sx¯ = √s n Where: sx¯ = standard error of the mean s = sample standard deviation n = sample size Equation 9.8: Single Sample t-Test for a Mean t = x¯ − µ0 √s n Where: t = t-statistic x¯ = sample mean µ0 = the population mean s = sample standard deviation n = sample size Equation 9.11: Single Sample z-Test for a Proportion z = pˆ − p0 σpˆ Where: z = z-statistic pˆ = sample proportion p0 = the population proportion σpˆ = the estimated standard error of the proportion Equation 9.12: Estimated Standard Error of the Proportion σpˆ = rp0(1 − p0) n Where: σpˆ = the estimated standard error of the proportion p0 = the population proportion n = sample size Equation 9.13: Single Sample z-Test for a Proportion z = qpˆ − p0 p0(1−p0) n Where: z = z-statistic pˆ = sample proportion p0 = the population proportion n = sample size Chapter 10 Estimation and Hypothesis Testing II: Independent and Paired Sample T-Test Learning Objectives: 1. Describe the difference between an independent and a dependent sample. 2. Describe the difference between and uses of an independent sample t-test and a paired sample t-test. 3. List the assumptions for the independent sample t-test. 4. Test a hypothesis using two independent samples. 5. Construct confidence intervals for the difference in means for the independent sample t- test. 6. List the assumptions for the paired sample t-test. 7. Test a hypothesis using two dependent samples. 8. Construct confidence intervals for the difference in means for the paired sample t-test. Chapter Summary In this chapter, students’ extend their learning about statistical hypothesis testing in relation to means. Specifically, students’ learn the processes and formulas involved in comparing two sample means (as opposed to one same mean compared to the population with known or unknown parameters). The distinction between independent and dependent samples is presented; as is how to compare means with these two distinct forms of data. Key Formulas The following represent the key formulas for this chapter. PowerPoint slides are provided for each chapter. In addition to these slides, a PDF file containing only the formulas are also provided. Independent sample t-test for two means Degrees of freedom for the independent sample t- test. Confidence interval for the independent sample t- test Paired sample t-test for two means Degrees of freedom for the paired sample t-test. Confidence interval for the paired sample t-test Interactive Figures: The textbook contains interactive figures. You may wish to use these in a lecture. Students also have access to these. For this chapter, there are two interactive figures. 1. Figure 10.2 demonstrates hypothesis testing for an independent sample t-test. 2. Figure 10.3 demonstrates hypothesis testing for an paired sample t-test. These interactive figures can be found on in the eBook and the Library under Chapter 10 Resources. Typical Lecture Material We have provided two sample lectures below. You may wish to add in additional discipline specific information to make these more relevant to your students. Lecture 1: Objective: To understand the distinctions between independent and paired-sample t-tests and their assumptions. Review the following concept table with your students; and help them to fill in the definitions of each statistical concept. The definitions that the students should come up with are in italics. Statistical Concept Defintion Independent Samples Samples are independent if they consist of diferent individuals and the selection of the individuals in the first group does not influence the selection of the second group. Dependent Samples Samples are dependent if they consis of the same individuals and/or the selection of the individuals in the first determines the selection of the second group. . Assumption of homogeneity of variances The assumption that the variances are equal across both groups. Violating an assumption The situation when our data does not meet one of the assumption(s) of our test. The pooled estimated standard error of the difference in the means. Is the estimated standard error used when the assumption of homogeneity of variances has been met. The pooled sample variance Is an unbiased estimator of the variance that both groups have in common. Difference between the paired scores The difference between the first and second score for each individual or pairs of individuals. Example 1: As students evolve in their statistical knowledge, it is important to know what type of data they have, what statistical test(s) are appropriate for that data; and what assumptions are being made when applying particular statistical tests. Ask your students to fill in the following information: (Answers are in italics) 1. What is the shared assumption between the independent samples t-test and the paired sample t-test? a. The variables must be measured at either the interval or ratio level of measurement. 2. What are the remaining three assumptions of the independent sample t-test? a. The two groups from which you collect the data must be independent of one- another. b. The variance in the population must be equal for both groups. c. The data must be normally distributed so that we can use the t-distribution. 3. What are the remaining two assumptions of the paired sample t-test? a. The two groups (and more specifically the data) must be dependent on one another. b. The differences between the two samples must be relatively normally distributed. Example 2: Ask your students to identify whether the following necessitates an independent sample t-test or a paired sample t-test. The correct response is in italics. 1. Does the mean level of academic achievement among the 4th grade students significantly change following the introduction of a teaching assistant in the classroom? (paired- sample t-test) 2. Does the mean level of physical activity significantly vary between males and females? (independent sample t-test) 3. Did the number of cigarettes smoked by the 100 research participants decrease after they went through the smoking reduction program? (paired-sample t-test) Example 3: Use the interactive figure for Figure 10.2 (shown below) to demonstrate how an independent sample t-test works. You can adjust the values to simultaneously estimate the t-statistic and test the null hypothesis that there is no difference in the mean between the first and second group. The numbers are coloured so that students can follow where they are coming from. Example 4: Use the interactive figure for Figure 10.3 (shown below) to demonstrate how a paired sample t- test works. You can adjust the values to simultaneously estimate the t-statistic and test the null hypothesis that there is no difference in the mean between Week 1 and Week 2. The numbers are coloured so that students can follow where they are coming from. Lecture 2: Objective: To be able to conduct an independent sample t-test hypothesis and a paired sample t- test hypothesis. Example 1: Read the following scenario to your students and ask them to complete the five steps of hypothesis testing (using an alpha value of 0.05), including computing the 95% confidence interval for the point estimate. You are interested in testing if there is a difference between grade 10 students from two different schools (School A and School B) in regards to their opinion on the following statement: “I think that Canadian politics are irrelevant to my life.” The responses of the students were captured on a 5-point scale ranging from 1 = Strongly Agree and 5 = Strongly Disagree. The data for the responses are as follows: School A: School B: 𝑥̅= 2.90 𝑥̅= 3.30 𝑠2 = 1.20 𝑠2 = 1.20 𝑛 = 80 𝑛 = 80 Note: You can use the interactive Figure 10.2 to demonstrate this example after the students have calculated it. Answer: 1. Define the null and alternative hypothesis H0 : μ1 = μ2 HA : μ1 ≠ μ2 2. Define the sampling distribution and critical values The mean of the sampling distribution is 0 (μ1 - μ2 = 0). The pooled estimated standard error of difference in the means is: 𝑠𝑥̅1− 𝑠𝑥̅2= √ (𝑛1 − 1)𝑠12 + (𝑛2 − 1)𝑠22 (𝑛1 + 𝑛2 − 2) × ( 1 𝑛1 + 1 𝑛2) 𝑠𝑥̅1− 𝑠𝑥̅2= √ (80 − 1)1.20 + (80 − 1)1.20 (80 + 80 − 2) × ( 1 80 + 1 80) 𝑠𝑥̅1− 𝑠𝑥̅2= √ (79)1.20 + (79)1.20 158 × (0.0125 + 0.0125) 𝑠𝑥̅1− 𝑠𝑥̅2= √94.80 + 94.80 158 × 0.025 𝑠𝑥̅1− 𝑠𝑥̅2= √189.60 158 × 0.025 𝑠𝑥̅1− 𝑠𝑥̅2= √1.20 × 0.025 = √0.03 = 0.17 The sampling distribution would look like the following: With 158 degrees of freedom (df = 80 + 80 – 2 = 158) and an alpha value of 0.05, the critical value of t is ± 1.96. Note: In the interactive, the critical value of t will show as ± 1.975 because it is calculating the exact value of t rather than using approximate values as is done with the t-table when n > 120. The remainder of this example is based on the critical value of t being ± 1.96. 3. Calculate the test-statistic using the sample data 𝑧 = 𝑥̅1− 𝑥̅2 𝑠𝑥̅1− 𝑠𝑥̅2= 2.90 − 3.30 0.17 = −0.40 0.17 = −2.35 4. Make the decision regarding the hypothesis Given that our t-statistic of -2.35, is outside of the + 1.96 area (-2.35 -2.15 < +2.262), we fail to reject the null hypothesis that 𝜇𝐷 ̅ = 0. Therefore, although there is a mean difference of -2.15 in this sample, it is not statistically significant at the alpha = 0.05 level. The 95% confidence interval of the difference between the paired scores is: 𝐶𝐼 = 𝐷 ̅ ± 𝑡(𝛼=0.05)(𝑠𝐷 ̅ ) 𝐶𝐼 = −2.30 ± 2.262(1.07) 𝐶𝐼 = −2.30 ± 2.42 Lower Limit of 𝐶𝐿 = −2.30 − 2.42 = −4.72 Upper Limit of 𝐶𝐿 = −2.30 + 2.42 = 0.12 Therefore the 95% confidence interval of the difference in the means is: −4.72 ≤ 𝜇𝐷 ̅ ≤ 0.12 5. Interpret the results Based on our sample, there appears to not be a statistically significant difference between the number of words recited correctly at Time 1 and Time 2. The 95% confidence interval of the difference is [-4.72, 0.12], which means that we are 95% confident that the difference in the means for the population is between -4.72 and 0.12. Solutions to End-of-Chapter Problems Problem 10-1 a) -6.89  1 - 2  4.69 (LO4) b) Computed statistic = -0.54; critical value for a two tailed test with 20 degrees of freedom = 2.845 Conclusion: the null hypotheses would be accepted. There is not sufficient evidence. (LO5) c) P-value ≥ 0.25 from table. Actual p-value for 20 degrees of freedom = 0.5952 (LO4/LO5) Problem 10-2 a) -5.77 1 - 2  9.57 (LO4) b) Computed statistic = 0.5075; critical value for a two tailed test with 28 degrees of freedom = 2.845 Conclusion: the null hypotheses would be accepted. There is not sufficient evidence. (LO5) c) P-value ≥ 0.25 from table. Actual p-value for 28 degrees of freedom = 0.6158 (LO4/LO5) Problem 10-3 a) -10.12  D  9.57 (LO8) b) Computed statistic = -2.677; critical value for a two tailed test with 7 degrees of freedom = 2.365 Conclusion: the null hypotheses would be rejected. There is sufficient evidence. (LO7) c) 0.025  P-value  0.05 from table. Actual p-value for 7 degrees of freedom = 0.0317 (LO7/LO8) Solutions to Interactive Exercises Question 10-1 Management of a sales company wants to determine whether or not a 3 day motivational retreat would increase the sales of their sales professionals. A sample of 8 sales people were selected and sent to the course. Their sales (in $ 1000s) before and after the retreat is as given below: ID Sales Before Sales After 1 43 48 2 23 22 3 36 32 4 18 23 5 27 32 6 32 34 7 42 49 8 36 54 Use α= 0.05, and test to see if it can be concluded that the retreat actually increased the sales. Feedback: Hypothesis test results: μD : mean of the paired difference between Before and After H0 : μD = 0 HA : μD< 0 Test Statistic = t = -2.0054 T critical = -1.895 As test statistic is less than the t critical value, we reject Ho. There is sufficient evidence to believe at 0.05 level of significance that the sales have increased after the retreat. Question 10-2 A psychologist is examining the influence of an older sibling in language development. She selected a random sample of 20 – 4 year old children with no siblings and another sample of 20 - 4 year old children with siblings were given a test on language skills. The score for each child was recorded and the mean and the standard deviation of the set is also noted. The data is given below: Group with No Siblings Group with Older Siblings n = 20 n = 20 Mean = 78 Mean = 89 SD = 12 SD = 16 Does this data indicate that having older siblings will improve language skills? Use α= 0.05 Feedback: μ1 : mean of population 1 (Group with NO siblings) μ2 : mean of population 2 (Group with OLDER siblings) μ1 - μ2 : mean difference H0 : μ1 - μ2 = 0 HA : μ1 - μ2 < 0 (with pooled variances) Test Statistic = -2.46 DF = 38 t Critical = -1.686 Test statistic is smaller than the critical value. Hence reject H0. There is sufficient evidence to believe that having older siblings will improve language skills. Question 10-3 Management of a sales company wants to determine whether or not a 3 day motivational retreat would increase the sales of their sales professionals. A sample of 8 sales people were selected and sent to the course. Their sales (in $ 1000s) before and after the retreat is as given below: ID Sales Before Sales After 1 43 48 2 23 22 3 36 32 4 18 23 5 27 32 6 32 34 7 42 49 8 36 54 Construct a two-sided 95% confidence interval and see whether there is any significant difference in the sales before and after the retreat. Feedback: (-10.0786, 0.8286) As Zero is in the interval, it can be concluded that there is NO significant difference in sales before and after the retreat. Solutions to SPSS Exercises Question 10-1 It is surmised that professional golfers play better towards the end of the golf season. The following data represents the first and last round of golf for a professional golfer. a b 73 66 68 70 73 65 71 77 67 67 68 66 76 70 66 71 Using a dependent sample t-test, does the data indicate that the mean score at the end of the season is higher than the mean score at the beginning of the season? Ans: Using SPSS analyze - compare means – dependent sample t-test we see the following result. Paired Samples Statistics Mean N Std. Deviation Std. Error Mean Pair 1 VAR00007 70.2500 8 3.53553 1.25000 VAR00008 69.0000 8 3.92792 1.38873 The remainder of the tables can be found in the output file for SPSS on Chapter 10. With a significance of 0.535 the null hypothesis is maintained and there is no difference between the first and last game of the season. LO: 7 Page: 282-288 Equations from Chapter 10 Scott R. Colwell and Edward M. Carter c 2012 Equation 10.1: Independent Sample t-Test for Two Means t = x¯1 − x¯2 sx¯1−x¯2 Where: t = t-statistic x¯1 = sample mean for group 1 x¯2 = sample mean for group 2 sx¯1−x¯2 = estimated standard error of the difference in the means Equation 10.2: Pooled Estimated Standard Error of the Difference of the Means sx¯1−x¯2 = s(n1 − 1)s 2 1 + (n2 − 1)s22 (n1 + n2 − 2) × 1 n1 + 1 n2 Where: sx¯1−x¯2 = estimated standard error of the difference in the means n1 = sample size for group 1 s12 = sample variance for group 1 n2 = sample size for group 2 s22 = sample variance for group 2 Equation 10.3: Pooled Sample Variance sp2 = (n1 − 1)s 2 1 + (n2 − 1)s22 (n1 + n2 − 2) Where: sp2 = pooled sample variance n1 = sample size for group 1 s12 = sample variance for group 1 n2 = sample size for group 2 s22 = sample variance for group 2 Equation 10.14: Expanded Formula for the Independent Sample t-Test t = r x¯1 − x¯2 (n1−1)s12+(n2−1)s22 (n1+n2−2) × 1 n1 + 1 n2 Where: t = t-statistic x¯1 = sample mean for group 1 x¯2 = sample mean for group 2 n1 = sample size for group 1 s12 = sample variance for group 1 n2 = sample size for group 2 s22 = sample variance for group 2 Equation 10.16: Degrees of Freedom for the Independent Sample t-Test df = n1 + n2 − 2 Where: df = degrees of freedom n1 = sample size for group 1 n2 = sample size for group 2 Equation 10.20: Confidence Interval of the Difference in Two Population Means CI = (x¯1 − x¯2) ± t(α)(sx¯) Where: CI = Confidence interval of the difference x¯1 = sample mean for group 1 x¯2 = sample mean for group 2 t(α) = t-score associated with the level of significance sx¯ = estimated standard error of the difference in the means (using pooled variances) Equation 10.23: Paired Sample t-Test t = D¯ sD¯ Where: t = t-statistic D¯ = mean difference between the paired scores sD¯ = estimated standard error of the difference between the paired scores Equation 10.24: Mean of the Difference Between the Paired Scores D¯ = Pd n Where: D¯ = mean difference between the paired scores d = individual differences between the paired scores n = sample size Equation 10.26: Estimated Standard Error of the Difference Between the Paired Scores sD¯ = √sd n Where: sD¯ = estimated standard error of the difference between the paired scores sd = sample standard deviation of the difference between the paired scores n = sample size Equation 10.27: Standard Deviation of the Difference Between the Paired Scores sd = sP(d − D¯)2 n − 1 Where: sd = sample standard deviation of the difference between the paired scores d = individual differences between the paired scores D¯ = mean difference between the paired scores n = sample size Equation 10.29: Expanded Formula for the Paired Sample t-Test t = D¯ √sdn Where: t = t-statistic D¯ = mean difference between the paired scores sd = sample standard deviation of the difference between the paired scores Equation 10.31: Degrees of Freedom for the Paired Sample t-Test df = n − 1 Where: df = degrees of freedom n = sample size Equation 10.34: Confidence Interval for the Mean Difference Between Two Paired Scores in the Population CI = D¯ ± t(α)(sD¯) Where: CI = Confidence interval of the difference D¯ = mean difference between the paired scores t(α) = t-score associated with the level of significance sD¯ = estimated standard error of the difference Solution Manual for Introduction to Statistics for Social Sciences Scott R. Colwell, Edward M. Carter 9780071319126