Section 9.3 Inferences About Two Means (Independent) Objective Compare the proportions of two independent means using two samples from each population. Hypothesis Tests and Confidence Intervals of two proportions use the t-distribution

Definitions Two samples are independent if the sample values selected from one population are not related to or somehow paired or matched with the sample values from the other population Examples: Flipping two coins (Independent) Drawing two cards (not independent) Text will use the wording ‘matched pairs’. Example at bottom of page 469-470 of Elementary Statistics, 10th Edition

Notation First Population μ1 First population mean σ1 First population standard deviation n1 First sample size x1 First sample mean s1 First sample standard deviation

Notation Second Population μ2 Second population mean σ2 Second population standard deviation n2 Second sample size x2 Second sample mean s2 Second sample standard deviation

Requirements (1) Have two independent random samples (2) σ1 and σ2 are unknown and no assumption is made about their equality (3) Either or both the following holds: Both sample sizes are large (n1>30, n2>30) or Both populations have normal distributions All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval

Tests for Two Independent Means The goal is to compare the two Means H0 : μ1 = μ2 H1 : μ1 ≠ μ2 H0 : μ1 = μ2 H1 : μ1 < μ2 H0 : μ1 = μ2 H1 : μ1 > μ2 Two tailed Left tailed Right tailed Example given at the bottom of page 458-460 of Elementary Statistics, 10th Edition. Note: We only test the relation between μ1 and μ2 (not the actual numerical values)

Finding the Test Statistic Note: m1 – m2 =0 according to H0 Degrees of freedom: df = smaller of n1 – 1 and n2 – 1. This equation is an altered form of the test statistic for a single mean when σ unknown (see Ch. 8-5)

Test Statistic Degrees of freedom df = min(n1 – 1, n2 – 1)   Degrees of freedom df = min(n1 – 1, n2 – 1) Example given at the bottom of page 458-460 of Elementary Statistics, 10th Edition. Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics)

Steps for Performing a Hypothesis Test on Two Independent Means Write what we know State H0 and H1 Draw a diagram Find the Test Statistic Find the Degrees of Freedom Find the Critical Value(s) State the Initial Conclusion and Final Conclusion Note: Same process as in Chapter 8

Example 1 A headline in USA Today proclaimed that “Men, women are equal talkers.” That headline referred to a study of the numbers of words that men and women spoke in a day. Use a 0.05 significance level to test the claim that men and women speak the same mean number of words in a day.

Example 1 H0 : µ1 = µ2 H1 : µ1 ≠ µ2 n1 = 186 n2 = 210 α = 0.05 x1 = 15668.5 x2 = 16215.0 Claim: μ1 = μ2 s1 = 8632.5 s2 = 7301.2 H0 : µ1 = µ2 H1 : µ1 ≠ µ2 Two-Tailed H0 = Claim t-dist. df = 185 t = 7.602 -tα/2 = -1.97 tα/2 = 1.97 Test Statistic   Degrees of Freedom df = min(n1 – 1, n2 – 1) = min(185, 209) = 185 Critical Value tα/2 = t0.025 = 1.97 (Using StatCrunch) Initial Conclusion: Since t is not in the critical region, accept H0 Final Conclusion: We accept the claim that men and women speak the same average number of words a day.

Stat → T statistics → Two sample → With summary Example 1 n1 = 186 n2 = 210 α = 0.05 x1 = 15668.5 x2 = 16215.0 Claim: μ1 = μ2 s1 = 8632.5 s2 = 7301.2 H0 : µ1 = µ2 H1 : µ1 ≠ µ2 Two-Tailed H0 = Claim Stat → T statistics → Two sample → With summary Null: prop. diff.= Alternative Sample 1: Mean Std. Dev. Size Sample 2: Mean ● Hypothesis Test 15668.5 Using StatCrunch 8632.5 186 ≠  16215.0 (Be sure to not use pooled variance) 7301.2 210 (No pooled variance) P-value = 0.4998 Initial Conclusion: Since P-value > α (0.05), accept H0 Final Conclusion: We accept the claim that men and women speak the same average number of words a day.

Confidence Interval Estimate We can observe how the two proportions relate by looking at the Confidence Interval Estimate of μ1–μ2 CI = ( (x1–x2) – E, (x1–x2) + E )   2 2 Example at the bottom of page 461 of Elementary Statistics, 10th Edition Rationale for the procedures of this section on page 462-463 of Elementary Statistics, 10th Edition Where df = min(n1–1, n2–1)

Example 2 Use the same sample data in Example 1 to construct a 95% Confidence Interval Estimate of the difference between the two population proportions (µ1–µ2) n1 = 186 n2 = 210 x1 = 15668.5 x2 = 16215.0 s1 = 8632.5 s2 = 7301.2 df = min(n1–1, n2–1) = min(185, 210) = 185 tα/2 = t0.05/2 = t0.025 = 1.973 x1 - x2 = 15668.5 – 16215.0 = -546.5 df = min(n1–1, n2–1) = min(185, 210) = 185 tα/2 = t0.1/2 = t0.05 = 1.973 x1 - x2 = 15668.5 – 16215.0 = -546.5 (x1 - x2) + E = -546.5 + 1596.17 = 1049.67 (x1 - x2) – E = -546.5 – 1596.17 = -2142.67   CI = (-2142.7, 1049.7)

Stat → T statistics → Two sample → With summary Example 2 Use the same sample data in Example 1 to construct a 95% Confidence Interval Estimate of the difference between the two population proportions (µ1–µ2) n1 = 186 n2 = 210 x1 = 15668.5 x2 = 16215.0 s1 = 8632.5 s2 = 7301.2 Stat → T statistics → Two sample → With summary Level: Sample 1: Mean Std. Dev. Size Sample 2: Mean ● Confidence Interval 15668.5 8632.5 0.95 186 16215.0 Using StatCrunch 7301.2 (No pooled variance) 210 CI = (-2137.4, 1044.4) Note: slightly different because of rounding errors

Example 3 Consider two different classes. The students in the first class are thought to generally be older than those in the second. The students’ ages for this semester are summed as follows: (a) Use a 0.1 significance level to test the claim that the average age of students in the first class is greater than the average age of students in the second class. (b) Construct a 90% confidence interval estimate of the difference in average ages. n1 = 93 n2 = 67 x1 = 21.2 x2 = 19.8 s1 = 2.42 s2 = 4.77

Example 3a H0 : µ1 = µ2 H1 : µ1 > µ2 n1 = 93 n2 = 67 α = 0.1 x1 = 21.2 x2 = 19.8 Claim: µ1 > µ2 s1 = 2.42 s2 = 4.77 H0 : µ1 = µ2 H1 : µ1 > µ2 Right-Tailed H1 = Claim t-dist. df = 66 Test Statistic tα/2 = 1.668 t = 7.602   Degrees of Freedom df = min(n1 – 1, n2 – 1) = min(92, 66) = 66 Critical Value tα/2 = t0.05 = 1.668 (Using StatCrunch) Initial Conclusion: Since t is in the critical region, reject H0 Final Conclusion: We accept the claim that the average age of students in the first class is greater than that in the second.

Stat → T statistics → Two sample → With summary Example 3a n1 = 93 n2 = 67 α = 0.1 x1 = 21.2 x2 = 19.8 Claim: µ1 > µ2 s1 = 2.42 s2 = 4.77 H0 : µ1 = µ2 H1 : µ1 > µ2 Right-Tailed H1 = Claim Stat → T statistics → Two sample → With summary Null: prop. diff.= Alternative Sample 1: Mean Std. Dev. Size Sample 2: Mean ● Hypothesis Test 21.2 Using StatCrunch 2.42 93 ≠  19.8 (Be sure to not use pooled variance) 4.77 67 (No pooled variance) P-value = 0.0299 Initial Conclusion: Since P-value < α (0.1), reject H0 Final Conclusion: We accept the claim that the average age of students in the first class is greater than that in the second.

Example 3b mp CI = (0.34, 2.46) n1 = 93 n2 = 67 α = 0.1 x1 = 21.2 x2 = 19.8 s1 = 2.42 s2 = 4.77 (90% Confidence Interval) df = min(n1–1, n2–1) = min(92, 66) = 66 tα/2 = t0.1/2 = t0.05 = 1.668 x1 - x2 = 21.2 – 19.8 = 1.4 (x1 - x2) + E = 1.4 + 1.058 = 2.458 (x1 - x2) – E = 1.4 – 1.058 = 0.342   mp CI = (0.34, 2.46)

Stat → T statistics → Two sample → With summary Example 3b n1 = 93 n2 = 67 α = 0.1 x1 = 21.2 x2 = 19.8 s1 = 2.42 s2 = 4.77 (90% Confidence Interval) Stat → T statistics → Two sample → With summary Null: prop. diff.= Alternative Sample 1: Mean Std. Dev. Size Sample 2: Mean ● Hypothesis Test 21.2 Using StatCrunch 2.42 93 ≠  19.8 (Be sure to not use pooled variance) 4.77 67 (No pooled variance) CI = (0.35, 2.45)