CH-7-8 The Foundations of Inferential Statistics: The Scientific Method and Hypotheses Testing – Introduction to Statistics for Social Sciences : Book Guide

This Document Contains Chapters 7 to 8 Chapter 7 The Foundations of Inferential Statistics: The Scientific Method and Hypotheses Testing Learning Objectives: 1. Explain the importance of the scientific method. 2. Describe the difference between a null and alternative hypothesis. 3. Write hypotheses including both the null and the alternative. 4. Explain the difference between one-tailed and two-tailed hypotheses. 5. Describe the importance of significance levels and critical values. 6. Explain what it means to reject and fail to reject the null hypothesis. 7. Describe the five steps of hypothesis testing. 8. Explain Type I and Type II errors. Chapter Summary In this chapter, students were introduced to the process of hypothesis testing as it relates to statistical inference. The chapter begins with a discussion about the scientific method, followed by a description of the concepts of hypothesis and significance; and then ties these concepts together with a discussion about critical values. The chapter ends by outlining the five steps involved in hypothesis testing and the distinction between Type I and Type II error. 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. Standardized Score for sampling distribution 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. Figure 7.9 illustrates a two-tailed hypothesis test. These interactive figures can be found on in the eBook and the Library under Chapter 7 Resources. This Document Contains Chapters 7 to 8 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: Understand the concept of the scientific method and hypothesis testing, the distinction between a one-tailed and two-tailed hypothesis; and the use of critical values and significant levels. 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 Population Parameter Estimation The analytical method for calculating the expected population parameter based on a sample statistic. Statistical Hypothesis Testing The use of statistical methods to test a hypothesis. The Scientific Method The process by which we investigate phenomena whether natural or social in nature. Statistically Signficant When a result of our study is unlikely to have occurred due to chance. Null Hypothesis States that the result was due to chance (from sampling error) and as such is not a true difference. Alternative Hypothesis States that the result was not due to chance (from sampling error) and represents a real difference or effect or relationship. One-Tailed Hypothesis A hypothesis that states a specific direction. Two-Tailed Hypothesis A hypothesis that does not state a specific direction. Siginifciance Level Is a value that is used to determine whether to reject or fail to reject the null hypothesis. Critical Value Is the corresponding value to the significance level that determines the boundry for rejecting or failing to reject the null hypothesis. Test Statistic A summary value (statistic) that is based on our sample data and tells us how far our sample estimate (eg. Mean) differs from zero under the null hypothesis distribution. Type I Error The probability that you will reject the null hypothesis when it is actually true. Type II Error The probability that you will fail to reject a false null hypothesis. Example 1: Read the following scenario to your students: You are interested in studying gender differences in the occurrence of workplace harassment. Suppose you hypothesize that females experience more workplace harassment than males. Write the Null and Alternative Hypothesis in both words and in notation. Answer: H0: Female occurrence of workplace harassment is less than or equal to male occurrence of workplace harassment. HA: Female occurrence of workplace harassment is greater than male occurrence of workplace harassment. H0: μA ≤ μ0 HA: μA > μ0 Example 2: The following hypotheses are related to difference in perception of ethnic/racial discrimination in Canada. Ask your students to identify if each of the following alternative hypotheses are one- tailed or two-tailed. 1. Individuals aged 50 years and older perceive less ethnic/racial discrimination in Canada than younger individuals. Answer: one-tailed 2. Immigration status is related to perceptions of ethnic/racial discrimination in Canada Answer: two-tailed. 3. Ethnic minority status is positively related to perceived levels of ethnic/racial discrimination in Canada. Answer: one-tailed. 4. Those having a previous criminal record perceive different levels of ethnic/racial discrimination in Canada. Answer: two-tailed. 5. Those with greater levels of education perceive a greater amount of ethnic/racial discrimination in Canada. Answer: one-tailed. Example 3: Remind your students that a critical value is the corresponding value to the significance level that determines the boundary for rejecting or failing to reject the null hypothesis. Read the following to your students: Following the scientific method, you have used social segregation theory to inform your study on ethnic minority status and perceived levels of racial/ethnic discrimination in Canada. You and co-investigator differ on two important things, the hypothesis and the significance (alpha) level. The following represents researcher 1 (you) and researcher 2’s (your co-investigator) alternative hypotheses and chosen alpha level. The database you are using to test your hypothesis includes a sample of 1,000 males and 1,000 females. Researcher1: Females perceive different levels of ethnic/racial discrimination in Canada. (α = .05) Researcher2: Females perceive greater levels of ethnic/racial discrimination in Canada. (α = .01) 1. Plot a normal standard distribution, using both the alpha value and its associated critical value for each researcher’s hypothesis. 2. Identify the area under the curve where you would reject the null hypothesis and fail to reject the null hypothesis (i.e., Accept the alternative). Answer for Researcher 1 Answer for Researcher 2 Null hypothesis is two-tailed The shaded areas in both the left and right tails are where we reject the null hypothesis. The area in between the two shaded tails is where we fail to reject the null hypothesis. Null hypothesis is one-tailed The shaded area in the right tail is where we reject the null hypothesis. The area to the left of the shaded tail is where we fail to reject the null hypothesis. Lecture 2: Objective: To learn the five steps of hypothesis testing and their application; and to learn the concepts and interpretation of Type I and Type II error. Example 1: Note: the point of this chapter and exercise is for the student to understand how to test a hypothesis. Therefore this exercise focuses on the steps for hypothesis testing rather than the calculation of the test statistics. Read the following scenario to your students: Youth crime has become a significant concern within the school system. Of particular concern is the amount of school these youth tend to be missing; which could potentially put them at risk of failing their grade and potentially, early drop-out. Suppose that the mean (µ) number of missed school days among 150,000 youth between the age of 12-18 years old (the population of those registered in a provincial public school system) is 30 with a standard deviation (σ) of 10.12. Further suppose that researchers are currently studying whether students living in families with lower incomes differ in the number of missed school days over the course of the school year. Not knowing whether youth 12-18 with lower family incomes will miss a greater or lower number of school days than the population, the researcher’s hypothesize—assuming an alpha level of .01— that: HA : The number of missed school days among youth aged 12-18 who live in families with lower incomes will differ from the population mean. After collecting data from a random 2,000 students (who live in families with lower incomes), the researchers find that the mean (x) number of missed school days among these youth is 40 days. Write the five steps of hypothesis testing, describing what should be done at each step. Answer: Step 1: Define the Null and Alternative Hypothesis H0: The number of missed school days among youth aged 12-18 who live in families with lower incomes will not differ from the population mean. HA : The number of missed school days among youth aged 12-18 who live in families with lower incomes will differ from the population mean. This is a two-tailed hypothesis. Step 2: Define the Sampling Distribution and the Critical Values The sampling distribution would have a mean of zero and standard errors that represent how far potential values deviate from the mean. Given an alpha value of 0.01 and a two-tailed hypothesis, the critical values would be -2.576 and + 2.576 Step 3: Calculate the Test Statistic Using the Sample Data Using the appropriate test, which in this case is a single sample z-test, we calculate the test statistic. Step 4: Make the Decision Regarding the Hypothesis We now compare our test statistic to our critical value. If the test statistic is greater than -2.576 but less than +2.576 then we will fail to reject the null hypothesis. If the test statistic is less than -2.576 or greater than +2.576 then we will reject the null hypothesis. Step 5: Interpret the Results Once we know whether we have failed to reject or rejected the null hypothesis, we can make a statement regarding the results of our hypothesis test. Example 2: Provide your students with the following null and alternative hypothesis and then based on the provided scenarios, ask your students to identify which type of statistical error has occurred. H0: The number of missed school days among youth aged 12-18 who live in families with lower incomes will not differ from the population mean. HA : The number of missed school days among youth aged 12-18 who live in families with lower incomes will differ from the population mean. 1. The conclusion you make from your study is that the null hypothesis should be rejected. Thus, you conclude that there is a difference between the number of missed school days among youth who live in families with lower incomes and the population mean. However, the null was actually true, meaning there isn’t a difference. Is this a Type I or Type II error? Answer: Type I error. When we reject a null hypothesis that is actually true, we commit a Type I error. 2. The conclusion you make from your study is that the null hypothesis should not be rejected. Thus, you fail to reject the null hypothesis and conclude that there is no difference between the number of missed school days among youth who live in families with lower incomes and the population mean. However, the null was actually false, meaning there is a difference. Answer: Type II error. When we fail to reject a null hypothesis that is actually false, we commit a Type II error. Solutions to End-of-Chapter Problems Problem 7-1 a. Null ≥75; alternative 2500000 (LO3/LO4) Problem 7-2 a. Sample mean = 11.49; sample standard deviation = 0.47. (LO5/LO6) b. With Null =11 alternative ≠11, the p-value is less than . (LO5/LO6) Problem 7-3 With Null ≤24 and alternative >24,the p-value = 0.0179 >  so the claim is supported. (LO5/LO6) Solutions to Interactive Exercises Question 7-1 Sodium is needed in the body to regulate fluids and blood pressure smoothly. The accepted intake of sodium for children aged 4 to 8 is 1200 mg, with a standard deviation of 300 mg. A nutritionist believes that these days children consume more sodium than the accepted level. She took a sample of 25 children and found that the mean amount of sodium consumed is 1284. Is there significant evidence to support the researcher’s claim? (Assume that all the required conditions are satisfied in this test. Also take α= 0.05) Feedback: Let µ be the average sodium consumption of the children. H0 : µ= 1200 HA : µ>1200, Test Statistic = Z = 1.4 (Use equation 7.2) P value = P( Z>1.4) = 0.0681. P value > α, hence the test statistic was not significant. There isn’t enough evidence to believe that the sodium consumption is more than 1200 mg. Question 7-2 To test H0 : µ= 75 against HA : µ5.0, a simple random sample of size n = 36 is obtained from a population that is normally distributed. a) Does the population have to be normally distributed to test this hypothesis? Why? b) Assuming all the conditions satisfied, what is the test statistic if sample mean = 6.0. The population standard deviation is 1.5. c) What is the P-value? d) What is your decision? Feedback: a) The sample size is large (n>30), hence it is not necessary to have a normally distributed population. (Central Limit Theorem) b) Z = 4.0 c) P value = P (Z > 4.0) = almost zero. d) Null hypothesis is rejected. The test statistic is significant. Solutions to SPSS Exercises Question 7-1 A random sample of 14 brokerage houses on Bay Street gave the following price/earnings ratios. 24 16 22 15 19 22 23 14 12 13 17 13 11 18 The sample mean is 17.1. Assume that x is normally distributed with mean = 19 and standard deviation 4.36. Does this data indicate that the price /earnings ratios are significantly different from 19 (use SPSS to determine the solution)? Since the given population mean is the same as the sample mean we can use the t-test function in SPSS as there is not a Z-test function for this. As such we see the following. t df Sig. (2-tailed) peratio -1.656 13 .122 In this case, the test statistic is less than 1.96 in absolute value. As such the mean is not significantly different from 19. LO: 7 Page: 196 Equations from Chapter 7 Scott R. Colwell and Edward M. Carter c 2012 Equation 7.2: Calculating the Test Statistic z = x¯ − µ0 σx¯ = x¯ − µ0 (σ ÷ √n) Where: x¯ = sample mean µ0 = population mean of the null hypothesis σx¯ = standard error of the mean σ = population standard deviation n = sample size Example of a Two-Tailed Hypothesis H0 : µA = µ0 HA : µA 6= µ0 Where: µA = population mean for the alternative hypothesis µ0 = population mean for the null hypothesis Example of a One-Tailed Hypothesis H0 : µA ≤ µ0 HA : µA > µ0 Where: µA = population mean for the alternative hypothesis µ0 = population mean for the null hypothesis Chapter 8 Parameter Estimation Using Confidence Intervals Learning Objectives: 1. Describe the concept and use of confidence intervals. 2. Calculate the confidence interval of a mean when the population standard deviation is known. 3. Calculate the confidence interval of a mean when the population standard deviation is unknown. 4. Calculate the confidence interval of a proportion. 5. Describe the Student’s t-distribution. 6. Explain when and why the t-distribution is used. 7. Interpret confidence intervals. Chapter Summary In this chapter, students continue their discussion and learning about parameter estimation; and specifically, this chapter focuses on the estimation of confidence intervals. The chapter starts with a general discussion about the concept of confidence and the equations for confidence intervals surrounding mean and proportion estimates. It then covers the process of calculating confidence intervals for the mean when the population standard deviation is known and unknown. Following that, the calculation of the confidence interval for a proportion when the population proportion is unknown is discussed. The latter includes a necessary discussion about the Student’s T-Distribution, the estimated standard error of the mean, and degrees of freedom. 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. Confidence interval of the mean when σ is known Standard error of the mean Estimated standard error of the mean Confidence interval of the mean when σ is unknown Standard error of the proportion Estimated standard error of the proportion Confidence interval of the proportion when p is unknown 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 8.4 illustrates the confidence interval of the mean. 2. Figure 8.7 provides an interactive comparison of the t-distribution and the standard normal distribution. These interactive figures can be found on in the eBook and the Library under Chapter 8 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: Understand the concept of confidence intervals, the needed information in order to estimate it and how it is calculated. 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 Proportion Also referred to as a percentage Confidence Interval Is a range of values that we expect will contain the true population parameter. Level of Confidence The estimated probability that the true value of the population parameter falls within the stated confidence interval. Standard Error of the Mean The amount that individual sample means vary from the population mean. Estimated Standard Error of the Mean Is the estimate of the standard error of the mean based on the sample standard deviation. Degrees of Freedom Are the number of independent observations/pieces of informaton used in estimating a parameter. Single sample confidence intervals, degrees of freedom are calculated as n-1. Standard Error of the Proportion The amount that individual sample proportions vary from the population proportion. Estimated Standard Error of the Proportion Is the estimate of the standard error of the proportion based on the sample proportion. Example 1: 1. Ask your students to identify which distribution we use for the following confidence interval calculations. A: Confidence interval of the mean when sigma is known. Answer: Standard normal distribution. B: Confidence interval of the mean when sigma is unknown. Answer: Student’s t-distribution C: Confidence interval of the proportion when sigma is unknown. Answer: Standard normal distribution. 2. Ask your students to identify what pieces of information necessary to estimate: A: The confidence interval of the mean when the population standard deviation is unknown: Answer: 1. The sample mean 2. The desired level of alpha 3. The t-value that corresponds with the desired level of alpha 4. The sample standard deviation 5. The sample size B: The confidence interval of the mean when the population standard deviation is known: Answer: 1. The sample mean 2. The desired level of alpha 3. The z-value that corresponds with the desired level of alpha 4. The population standard deviation 5. The sample size C: The confidence interval of the proportion when the population proportion is unknown: Answer: 1. The sample proportion 2. The desired level of alpha 3. The z-value that corresponds with the desired level of alpha 4. The sample size Example 2: Given that in practice we rarely know the population standard deviation of the mean, we have provided two practice examples (below) for your lectures based on σ being unknown. Practice Example 1: Read the following scenario to your students: Suppose we randomly select 71 individuals who had just completed their weekly grocery shopping. We find that the mean grocery bill is $275 with a standard deviation of $35. What would be the 95% confidence interval of the mean? Answer: The formula for the confidence interval is: CI = 𝑥̅± 𝑡(𝑎= .05) ( 𝑠 √𝑛 ) The critical value of t at n – 1 degrees of freedom (71 – 1 = 70) is 1.994 The estimated standard error of the mean is ( 𝑠 √𝑛 = 35 √71 = 35 8.43 = 4.15) The formula for the confidence interval then becomes: CI = 275 ± 1.994(4.15) The lower limit of the confidence interval is therefore: 275 − 8.28 = $266.7 The upper limit of the confidence interval is therefore: 275 + 8.28 = $283.3 We can then conclude that the 95% CI is: $266.7 ≤ μ ≤ $283.3 Practice Example 2: Read the following scenario to your students: We randomly select 150 university students across Canada and ask each one how many resumes they send out when looking for a summer job. The mean number of resumes sent out was 8 with a standard deviation of 1.6. What would be the 97.5% confidence interval of the mean? Answer: The formula for the confidence interval is: CI = 𝑥̅± 𝑡(𝑎= .025) ( 𝑠 √𝑛 ) The critical value of t at n – 1 degrees of freedom (150 – 1 = 149) is 1.96 The estimated standard error of the mean is ( 𝑠 √𝑛 = 1.6 √150 = 1.6 12.25 = 0.13) The formula for the confidence interval then becomes: CI = 8 ± 1.96(0.13) The lower limit of the confidence interval is therefore: 8 − 0.25 = 7.75 The upper limit of the confidence interval is therefore: 8 + 0.25 = 8.25 We can then conclude that the 97.5% CI is: 7.75 ≤ μ ≤ 8.25 Lecture 2: Objective: To understand how to calculate a confidence interval for a proportion when the population proportion is unknown. Practice Example 1: Read the following scenario to your students: Suppose we randomly select 85 students and ask them if they have a part-time job during the regular school year. We find that 65% of the sample indicate that they do. What would be the 99% confidence interval of the proportion? Answer: The formula for the confidence interval is: CI = 𝑝̂± 𝑧(𝑎= .01) (√𝑝 ̂(1−𝑝 ̂) 𝑛 ) The critical value of z is 2.576 The estimated standard error of the proportion is √𝑝̂(1 − 𝑝̂) 𝑛 = √ . 65(1 − .65) 85 = √ . 65 × .35 85 = √ . 23 85 = 0.052 The formula for the confidence interval then becomes: CI = 0.65 ± 2.576(0.052) The lower limit of the confidence interval is therefore: 0.65 − 0.13 = 0.52 The upper limit of the confidence interval is therefore: 0.65 + 0.13 = 0.78 We can then conclude that the 99% CI is: 0.52 ≤ p ≤ 0.78 Practice Example 2: Read the following scenario to your students: Suppose we randomly surveyed 200 people living in New Brunswick and found that 34% of the sample had opted to receive the flu vaccination last year. What would be the 95% confidence interval of the proportion? Answer: The formula for the confidence interval is: CI = 𝑝̂± 𝑧(𝑎= .05) (√𝑝 ̂(1−𝑝 ̂) 𝑛 ) The critical value of z is 1.96 The estimated standard error of the proportion is √𝑝̂(1 − 𝑝̂) 𝑛 = √ . 34(1 − .34) 200 = √ . 34 × .66 200 = √ . 22 200 = 0.033 The formula for the confidence interval then becomes: CI = 0.34 ± 1.96(0.033) The lower limit of the confidence interval is therefore: 0.34 − 0.06 = 0.28 The upper limit of the confidence interval is therefore: 0.34 + 0.06 = 0.40 We can then conclude that the 95% CI is: 0.28 ≤ p ≤ 0.40 Solutions to End-of-Chapter Problems Problem 8-1 a) Mean = 11.49; standard deviation = 0.47. (LO2) b) 10.96 ≤  ≤ 12.02 (LO2) c) 11.04 ≤  ≤ 11.94 (LO2) d) 95% is the proportion of samples that would compute an interval containing the true mean. (LO7) Problem 8-2 a) In this setting, p is the population proportion of students that drive cars to school that are driving convertiles. (LO4) b) 0.05 ≤ p ≤ 0.12 (LO4) c) 0.03 ≤ p ≤ 0.14 (LO4) d) The interval widens as the confidence level gets larger. (LO7) Problem 8-3 84 ≤  ≤ 88 (LO2) Problem 8-4 a) The t-distribution should be used when  is unknown and n is less than 30.(LO6) b) 14.45 ≤  ≤ 37.01 (LO5) c) 6.45 ≤  ≤ 45.01 (LO5) Solutions to Interactive Exercises Question 8-1 A random sample of 25 students who took GMAT test preparation course from a coaching centre had an average score of 582 and a standard deviation of 15. a) What is the estimated standard error of mean? b) If the sample size is quadrupled, what will the new standard error be, if everything else remains the same? c) If you want to construct a confidence interval, which distribution would you refer to? Why? Feedback: (a) 3 (b) 1.5 (When sample size is quadrupled, the standard error is half) (c) t-distribution; As population standard deviation is not known Question 8-2 A simple random sample of size 16 is drawn from a population. The sample mean is found to be 40. The population standard deviation of mean is 4. (i) Construct a 95% CI for µ. (ii) If confidence level is increased to 99%, what will the new confidence interval be? Feedback: (i) 40+/-1.96*4/sqrt (16) 40+/-1.96 38.04 ≤ µ ≤ 41.96 (ii) 40+/-2.57 *(1) 37.43≤µ≤42.57 (As confidence level increases, the width of the confidence interval also increases) Question 8-3 A sociologist wishes to conduct a poll to estimate the percentage of Canadians who prefer the privatization of health care. He took a random sample of 500 people and noted that 47 favoured privatization of health care. (i) What is the estimate of the population proportion of those who prefer privatisation of health care? (ii) Construct an 80% confidence interval of the proportion of Canadians who prefers privatization. Feedback: (i) Sample proportion = 47/500 = 0.094 (ii) 0.094 +/- 1.28 * 0.0131 (0.094+/- 0.0168) (0.0772, 0.1108) Solutions to SPSS Exercises Question 8-1 a) Generate 10 samples in SPSS of size n=20 from a normal distribution with mean=25 and standard deviation=4. Hint: use the function transform – compute variable. Ans: You would see something like this: 1.00 21.53 22.16 25.28 26.47 29.83 22.56 29.83 20.89 24.96 16.82 2.00 26.40 22.93 28.46 25.20 17.15 23.60 27.50 22.67 29.57 18.09 3.00 32.74 26.78 22.28 25.32 21.76 28.00 15.92 22.82 16.41 22.32 4.00 35.06 29.66 28.98 18.92 27.07 29.54 26.16 22.35 29.10 21.40 5.00 23.53 28.69 22.72 33.42 22.43 21.34 24.50 23.02 31.04 21.28 6.00 21.00 32.79 17.35 26.08 18.32 22.37 30.84 31.57 24.31 22.10 7.00 18.25 24.40 17.29 29.51 23.53 27.40 28.72 26.04 23.14 27.66 8.00 22.60 28.82 30.89 23.68 22.31 29.15 30.87 23.70 26.61 26.03 9.00 23.57 19.52 20.23 37.02 19.26 27.11 21.92 21.88 18.29 31.94 10.00 19.66 22.43 20.37 26.34 18.88 26.08 25.34 27.44 17.73 27.19 11.00 20.52 26.30 28.80 30.46 31.27 22.18 20.04 16.61 26.58 32.73 12.00 21.65 24.73 25.41 27.60 31.23 23.40 29.30 19.77 26.71 19.28 13.00 23.86 26.99 27.55 20.29 25.69 23.97 22.44 23.44 29.17 27.80 14.00 20.80 18.23 23.15 30.13 25.65 27.49 19.36 23.98 26.94 31.74 15.00 24.80 27.66 25.71 21.11 18.66 25.51 14.67 22.02 22.72 22.32 16.00 19.61 28.88 30.31 23.36 26.38 22.75 27.26 24.93 24.67 26.92 17.00 20.74 27.44 30.09 31.30 28.55 28.23 17.91 33.49 21.81 27.02 18.00 23.97 21.80 29.05 25.50 30.26 21.32 17.87 23.37 26.35 31.88 19.00 22.34 24.56 23.42 24.72 28.40 29.22 16.74 18.53 28.15 16.80 20.00 24.25 16.66 31.07 26.26 19.69 18.81 21.31 33.12 24.43 26.03 LO: 3 Page: 222-224 b) Use analyze – compare means – one sample t-test to generate 10 confidence intervals for the mean. Ans: One-Sample Test Test Value = 0 t df Sig. (2- tailed) Mean Difference 95% Confidence Interval of the Difference Lower Upper sample 1 25.189 19 .000 23.34403 21.4043 25.2838 sample 2 27.220 19 .000 25.07166 23.1438 26.9995 sample 3 25.802 19 .000 25.42139 23.3592 27.4835 sample 4 26.963 19 .000 26.63467 24.5671 28.7022 sample 5 23.057 19 .000 24.31545 22.1082 26.5227 sample 6 35.399 19 .000 25.00156 23.5233 26.4798 sample 7 19.791 19 .000 23.42522 20.9479 25.9025 sample 8 24.225 19 .000 24.08213 22.0014 26.1628 sample 9 27.763 19 .000 24.93443 23.0546 26.8142 sample 10 21.803 19 .000 24.86863 22.4814 27.2559 LO: 3 Page: 222-224 c) How many of the intervals contain the true mean? Ans: In this case – all of them. Yours may be different. LO: 3 Page: 222-224 Equations from Chapter 8 Scott R. Colwell and Edward M. Carter c 2012 Equation 8.1: Confidence Interval of the Mean CI = x¯ ± z(α)(SE) Where: CI = confidence interval x¯ = sample mean z(α) = z-score associated with the level of significance SE = standard error of the mean Equation 8.2: Confidence Interval of the Proportion CI = pˆ ± z(α)(SE) Where: CI = confidence interval pˆ = sample proportion z(α) = z-score associated with the level of signficance SE = standard error of the proportion Equation 8.3: Level of Confidence Level of Confidence = 100 × (1 − α) Where: α = level of significance Equation 8.9: CI of the Mean When σ is Known CI = x¯ ± z(α) √σn Where: CI = confidence interval x¯ = sample mean z(α) = z-score associated with the level of significance σ = population standard deviation n = sample size Equation 8.14: Standard Error of the Mean σx¯ = √σ n Where: σx¯ = standard error of the mean σ = population standard deviation n = sample size Equation 8.15: Estimated Standard Error of the Mean sx¯ = √s n Where: sx¯ = estimated standard error of the mean s = sample standard deviation n = sample size Equation 8.16: CI of the Mean When σ is Unknown CI = x¯ ± t(α)(sx¯) Where: CI = confidence interval x¯ = sample mean t(α) = t-score associated with the level of significance sx¯ = estimated standard error of the mean Equation 8.16 (expanded): CI of the Mean When σ is Unknown CI = x¯ ± t(α) √sn Where: CI = confidence interval x¯ = sample mean t(α) = t-score associated with the level of significance s = sample standard deviation n = sample size Equation 8.23: CI of the Proportion When p is Unknown CI = pˆ ± z(α)(SE) Where: CI = confidence interval pˆ = sample proportion z(α) = z-score associated with the level of significance SE = standard error of the proportion Equation 8.23 (expanded): CI of the Proportion When p is Unknown CI = pˆ ± z(α) rpˆ(1 − pˆ) n ! Where: CI = confidence interval pˆ = sample proportion z(α) = z-score associated with the level of significance n = sample size Equation 8.27: Standard Error of the Proportion σpˆ = rp(1 − p) n Where: σpˆ = standard error of the proportion p = population proportion n = sample size Equation 8.28: Estimated Standard Error of the Proportion spˆ = rpˆ(1 − pˆ) n Where: spˆ = estimated standard error of the proportion pˆ = sample proportion n = sample size Solution Manual for Introduction to Statistics for Social Sciences Scott R. Colwell, Edward M. Carter 9780071319126