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Section 9.3 Inferences About Two Means (Independent)

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.

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Section 9.3 Inferences About Two Means (Independent)

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  1. Section 9.3Inferences 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

  2. 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)

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

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

  5. (1) Have two independentrandom 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 Requirements All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval

  6. 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 Note: We only test the relation between μ1 and μ2 (not the actual numerical values)

  7. 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)

  8. Test Statistic Degrees of freedomdf= min(n1 – 1, n2 – 1) Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics)

  9. 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

  10. 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.

  11. n1 =186 n2 =210 α=0.05 x1 =15668.5 x2 =16215.0 Claim:μ1 =μ2 s1 =8632.5 s2 =7301.2 Example 1 H0:µ1 =µ2 H1:µ1 ≠µ2 t-dist. df = 185 Two-Tailed H0 =Claim 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.

  12. Stat → T statistics → Two sample → With summary n1 =186 n2 =210 α=0.05 x1 =15668.5 x2 =16215.0 Claim:μ1 =μ2 s1 =8632.5 s2 =7301.2 Example 1 Sample 1: Mean Std. Dev. Size Sample 2: Mean Std. Dev. Size ●Hypothesis Test H0:µ1 =µ2 H1:µ1 ≠µ2 Null: prop. diff.= Alternative Two-Tailed H0 =Claim 15668.5 8632.5 Using StatCrunch 0 186 ≠  16215.0 (Be sure to not use pooled variance) 7301.2 (No pooled variance) 210 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.

  13. 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 Where df = min(n1–1, n2–1)

  14. 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)

  15. 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) Sample 1: Mean Std. Dev. Size Sample 2: Mean Std. Dev. Size ●Confidence Interval Level: n1 =186 n2 =210 x1 =15668.5 x2 =16215.0 s1 =8632.5 s2 =7301.2 15668.5 8632.5 0.95 186 16215.0 7301.2 Using StatCrunch (No pooled variance) 210 CI = (-2137.4, 1044.4) Note: slightly different because of rounding errors

  16. 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 =93n2 =67 x1 =21.2x2 =19.8 s1 =2.42s2 =4.77

  17. n1 =93 n2 =67 α = 0.1 x1 =21.2 x2 =19.8 Claim: µ1 >µ2 s1 =2.42 s2 =4.77 Example 3a H0:µ1 =µ2 H1:µ1 >µ2 t-dist. df = 66 Right-Tailed H1 =Claim 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.

  18. Stat → T statistics → Two sample → With summary n1 =93 n2 =67 α = 0.1 x1 =21.2 x2 =19.8 Claim: µ1 >µ2 s1 =2.42 s2 =4.77 Example 3a Sample 1: Mean Std. Dev. Size Sample 2: Mean Std. Dev. Size ●Hypothesis Test H0:µ1 =µ2 H1:µ1 >µ2 Null: prop. diff.= Alternative Right-Tailed H1 =Claim 21.2 2.42 Using StatCrunch 0 93 ≠  19.8 (Be sure to not use pooled variance) 4.77 (No pooled variance) 67 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.

  19. n1 =93 n2 =67 α = 0.1 x1 =21.2 x2 =19.8 s1 =2.42 s2 =4.77 Example 3b (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)

  20. Stat → T statistics → Two sample → With summary n1 =93 n2 =67 α = 0.1 x1 =21.2 x2 =19.8 s1 =2.42 s2 =4.77 Example 3b Sample 1: Mean Std. Dev. Size Sample 2: Mean Std. Dev. Size ●Hypothesis Test (90% Confidence Interval) Null: prop. diff.= Alternative 21.2 2.42 Using StatCrunch 0 93 ≠  19.8 (Be sure to not use pooled variance) 4.77 (No pooled variance) 67 CI = (0.35, 2.45)

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