Review of Chapter 10 Comparing Two Population Parameters Lesson 10 - R Review of Chapter 10 Comparing Two Population Parameters

Objectives Identify the conditions that need to be satisfied in order to do inference for comparing two population means. Construct a confidence interval for the difference between two population means. Perform a significance test for the difference between two population means. Identify the conditions that need to be satisfied in order to do inference for comparing two population proportions. Construct a confidence interval for the difference between two population proportions. Perform a significance test for the difference between two population proportions.

Vocabulary None new

Conditions for Comparing 2 Means SRS Two SRS’s from two distinct populations Measure same variable from both populations Independence Samples are independent of each other (not the test for match pair designs) Ni ≥ 10ni Normality Both populations are Normally distributed In practice, (large sample sizes for CLT to apply) similar shapes with no strong outliers If all else fails then check graphs

t-Test Statistic Since H0 assumes that the two population means are the same, our test statistic is reduce to: Similar in form to all of our other test statistics (x1 – x2) t0 = ------------------------------- s12 s22 ----- + ----- n1 n2 Test Statistic:

Confidence Intervals Lower Bound: s12 s22 ----- + ----- Upper Bound: tα/2 is determined using the smaller of n1 -1 or n2 -1 degrees of freedom x1 and x2 are the means of the two samples s1 and s2 are the standard deviations of the two samples Note: The two populations need to be normally distributed or the sample sizes large s12 s22 ----- + ----- n1 n2 (x1 – x2) – tα/2 · PE ± MOE s12 s22 ----- + ----- n1 n2 (x1 – x2) + tα/2 ·

2-Proportion Requirements Testing a claim regarding the confidence interval of the difference of two proportions SRS - Samples are independently obtained using SRS (simple random sampling) Independence: n1 ≤ 0.10N1 and n2 ≤ 0.10N2; Normality: n1p1 ≥ 10 and n1(1-p1) ≥ 10 n2p2 ≥ 10 and n2(1-p2) ≥ 10

P-Value is the area highlighted Reject null hypothesis, if Classical and P-Value Approach – Two Proportions P-Value is the area highlighted Remember to add the areas in the two-tailed! -zα -zα/2 zα/2 zα z0 -|z0| |z0| z0 Critical Region p1 – p2 z0 = --------------------------------- p (1- p) 1 1 --- + --- n1 n2 where x1 + x2 p = ------------ n1 + n2 Test Statistic: Reject null hypothesis, if P-value < α Left-Tailed Two-Tailed Right-Tailed z0 < - zα z0 < - zα/2 or z0 > zα/2 z0 > zα

Confidence Interval – Difference in Two Proportions Lower Bound: Upper Bound: p1 and p2 are the sample proportions of the two samples Note: the same requirements hold as for the hypothesis testing (p1 – p2) – zα/2 · p1(1 – p1) p2(1 – p2) --------------- + -------------- n1 n2 (p1 – p2) + zα/2 · p1(1 – p1) p2(1 – p2) --------------- + -------------- n1 n2

Sample Size for Estimating p1 – p2 The sample size required to obtain a (1 – α) * 100% confidence interval with a margin of error E is given by zα/2 n = n1= n2 = p1(1 – p1) + p2(1 – p2) ------ E 2 rounded up to the next integer. If a prior estimates of pi are unavailable, the sample required is zα/2 n = n1= n2 = 0.5 ------ E 2 rounded up to the next integer, where pi is a prior estimate of pi. The margin of error should always be expressed as a decimal when using either of these formulas.

Summary and Homework Summary Homework Use inference and confidence interval toolkits to solve two population parameter problems H0 assumption of equal parameters simply test statistics Never use pooled standard deviations in t-test Use combined (not pooled) proportions in two proportion z-test Homework none

Review Questions True or False. Populations that are normally distributed or large sample sizes (n≥30) are usually a requirement in constructing confidence limits. True or False. Tests regarding matched pair differences are independent. True or False. Confidence intervals are usually in the form of a point estimate +/- margin of error. Requirements to test the difference between two matched paired population means include all but which one? a. Samples Independent b. n > 30 c. Simple random sample d. normally distributed

Two Sample vs Difference Match the following Types of Sampling Methods: Dependent Individuals selected in one sample determine those in 2nd Independent Individuals selected in one sample do not dictate those in 2nd

Example 1 In 2005 Gallup conducted a poll that surveyed 735 adults and 526 responded yes to a question about if prayer could heal. Back in 2001, Gallup conducted a similar survey and 485 of the 735 individuals surveyed answered yes to the same question. Test the claim that the proportion of adults who believe that prayer can heal has increased since 2001, at α = 0.10 level of significance.

Workout – Example 1 a. H0 : b. H1 : c. Conditions d. Test Statistic: e. Conclusion: p1 = p2 p1 = % of people believing in prayer 2005 p2 = % of people believing in prayer 2001 p1 > p2 SRS: assume survey represents an SRS Independence: more than 7,350 adults in US; assume independent from each other Normality: np1 = 526 np2 = 485 both > 10 n(1-p1) = 209 n(1-p2) = 250 both > 10 p1 – p2 2 prop-Ztest: z = -------------------------------- = 2.308 p-value = 0.0105 √(pc(1-pc)(1/n1 + 1/n2)) Since p-value < α (.01 < .10) we reject the null hypothesis and conclude the more people believe in prayer in 2005 than in 2001

Example 2 McDonald’s executives want to experiment with redesigning the waiting lines in its restaurants from one line leading to four registers rather than four lines to separate registers. Their data analyst checks the normality of the data and comes up with the results to the right. Test the claim that the mean waiting time in the single line is less than in the 4 lines at α = 0.05 level of significance. 1 Line 4 Lines Mean 2.255 2.475 St Dev 0.574433 1.012488 Nr 20

Workout – Example 2 a. H0 : b. H1 : c. Conditions d. Test Statistic: e. Conclusion: μ1 = μ4 μ1 = mean customer wait time of 1 line μ4 = mean customer wait time of 4 lines μ1 < μ4 SRS: assume samples represents an SRS Independence: more than 200 McDonalds in US; easy to assume independent from each other Normality: ASSUME (analyst checked normality) x-bar1 – x-bar4 2 sample-t-test: t = -------------------------------- = -0.845 p-value = 0.2023 √[ (s1)2 /n1 + (s2)2 /n2 ] Since p-value > α (.20 > .05) we fail to reject the null hypothesis and conclude we have insufficient evidence to say one-line is quicker than four-lines

Example 3 In the problem 1 calculate the 90% confidence interval about the difference between 2005 – 2001 population proportions. Locate your data from problem 1 on the interval and interpret (your conclusion). Interval: [0.01609, .09547]  1.6% to 9.5% 0 .01 .02 .03 .04 .05 .06 .07 .08 .09 We are 90% confident that the true difference in proportion of people who believe in prayer from 2005 to 2001 lies between 1.6% to 9.5%. Since 0 is not in this interval we have evidence to say they are different.

Example 4 Construct a confidence interval for p1 – p2 at the 95% confidence level, given x1 = 804, n1 = 874, x2 = 892 and n2 = 954. Have to assume SRS and independence conditions are met. Normality: 804 and 892 (np) and 70 and 62 (n(1-p)) are all > 10 2-prop-Z-interval: Interval: [-0.0389, 0.00874]  -3.9% to 0.8% We are 95% confident that the true difference in proportion of p1 – p2 lies between -3.9% to 0.8%. Since 0 is in this interval we have insufficient evidence to say they are different.

Example 5 A sports performance analyst wants to analyze the difference between the proportion of males and females who play golf. What sample size should he use if he wishes the estimate to be within 2 percentage points with a 99% level of confidence, assuming that 1) he uses the 2003 estimates of 32.9% male and 19.7% female from Golf Digest? 2) he does not use any prior estimates? zα/2 n = n1= n2 = p1(1 – p1) + p2(1 – p2) ------ E 2 = [(.329)(.671)+(.197)(.803)] ()2 = [.37895](2.576/0.02)2 = 6286.57  6287 zα/2 n = n1= n2 = 0.5 ------ E 2 = 0.5 (2.576/0.02)2 = 8294.72  8295