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1 Objective Compare the proportions of two independent means using two samples from each population. Hypothesis Tests and Confidence Intervals of two proportions.

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Presentation on theme: "1 Objective Compare the proportions of two independent means using two samples from each population. Hypothesis Tests and Confidence Intervals of two proportions."— Presentation transcript:

1 1 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 Section 9.3 Inferences About Two Means (Independent)

2 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 3 Notation μ 1 First population mean σ 1 First population standard deviation n 1 First sample size x 1 First sample mean s 1 First sample standard deviation First Population

4 4 Notation μ 2 Second population mean σ 2 Second population standard deviation n 2 Second sample size x 2 Second sample mean s 2 Second sample standard deviation Second Population

5 5 (1)Have two independent random samples (2)σ 1 and σ 2 are unknown and no assumption is made about their equality (3)Either or both the following holds: Both sample sizes are large (n 1 >30, n 2 >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 6 Tests for Two Independent Means The goal is to compare the two Means H 0 : μ 1 = μ 2 H 1 : μ 1 ≠ μ 2 Two tailedLeft tailedRight tailed Note: We only test the relation between μ 1 and μ 2 (not the actual numerical values) H 0 : μ 1 = μ 2 H 1 : μ 1 < μ 2 H 0 : μ 1 = μ 2 H 1 : μ 1 > μ 2

7 7 Note:  1 –  2 =0 according to H 0 Degrees of freedom: df = smaller of n 1 – 1 and n 2 – 1. Finding the Test Statistic This equation is an altered form of the test statistic for a single mean when σ unknown (see Ch. 8-5)

8 8 Test Statistic Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics) Degrees of freedom df = min(n 1 – 1, n 2 – 1)

9 9 Steps for Performing a Hypothesis Test on Two Independent Means Write what we know State H 0 and H 1 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 10 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

11 11 H 0 : µ 1 = µ 2 H 1 : µ 1 ≠ µ 2 Two-Tailed H 0 = Claim t = t α/2 = 1.97 t-dist. df = 185 Test Statistic Critical Value Initial Conclusion: Since t is not in the critical region, accept H 0 Final Conclusion: We accept the claim that men and women speak the same average number of words a day. -t α/2 = Example 1 n 1 = 186 n 2 = 210 α = 0.05 x 1 = x 2 = Claim: μ 1 = μ 2 s 1 = s 2 = Degrees of Freedom df = min(n 1 – 1, n 2 – 1) = min(185, 209) = 185 t α/2 = t = 1.97 (Using StatCrunch)

12 12 H 0 : µ 1 = µ 2 H 1 : µ 1 ≠ µ 2 Two-Tailed H 0 = Claim Initial Conclusion: Since P-value > α (0.05), accept H 0 Final Conclusion: We accept the claim that men and women speak the same average number of words a day. Example 1 n 1 = 186 n 2 = 210 α = 0.05 x 1 = x 2 = Claim: μ 1 = μ 2 s 1 = s 2 = Stat → T statistics → Two sample → With summary Null: prop. diff.= Alternative Sample 1: Mean Std. Dev. Size Sample 2: Mean Std. Dev. Size ● Hypothesis Test P-value = ≠ Using StatCrunch (No pooled variance) (Be sure to not use pooled variance)

13 13 Confidence Interval Estimate We can observe how the two proportions relate by looking at the Confidence Interval Estimate of μ 1 –μ 2 CI = ( (x 1 –x 2 ) – E, (x 1 –x 2 ) + E ) Where df = min(n 1 –1, n 2 –1) 22

14 14 df = min(n 1 –1, n 2 –1) = min(185, 210) = 185 t α/2 = t 0.1/2 = t 0.05 = x 1 - x 2 = – = (x 1 - x 2 ) + E = = (x 1 - x 2 ) – E = – = Use the same sample data in Example 1 to construct a 95% Confidence Interval Estimate of the difference between the two population proportions ( µ 1 –µ 2 ) Example 2 n 1 = 186 n 2 = 210 x 1 = x 2 = s 1 = s 2 = df = min(n 1 –1, n 2 –1) = min(185, 210) = 185 t α/2 = t 0.05/2 = t = x 1 - x 2 = – = CI = ( , )

15 15 Use the same sample data in Example 1 to construct a 95% Confidence Interval Estimate of the difference between the two population proportions ( µ 1 –µ 2 ) Example 2 CI = ( , ) n 1 = 186 n 2 = 210 x 1 = x 2 = s 1 = s 2 = Stat → T statistics → Two sample → With summary Level: Sample 1: Mean Std. Dev. Size Sample 2: Mean Std. Dev. Size ● Confidence Interval Using StatCrunch Note: slightly different because of rounding errors (No pooled variance)

16 16 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. Example 3 n 1 = 93 n 2 = 67 x 1 = 21.2 x 2 = 19.8 s 1 = 2.42 s 2 = 4.77

17 17 t = 7.602t α/2 = Test Statistic Critical Value Degrees of Freedom df = min(n 1 – 1, n 2 – 1) = min(92, 66) = 66 t α/2 = t 0.05 = (Using StatCrunch) H 0 : µ 1 = µ 2 H 1 : µ 1 > µ 2 Right-Tailed H 1 = Claim n 1 = 93 n 2 = 67α = 0.1 x 1 = 21.2 x 2 = 19.8Claim: µ 1 > µ 2 s 1 = 2.42 s 2 = 4.77 t-dist. df = 66 Example 3a Initial Conclusion: Since t is in the critical region, reject H 0 Final Conclusion: We accept the claim that the average age of students in the first class is greater than that in the second.

18 18 H 0 : µ 1 = µ 2 H 1 : µ 1 > µ 2 Right-Tailed H 1 = Claim Example 3a n 1 = 93 n 2 = 67α = 0.1 x 1 = 21.2 x 2 = 19.8Claim: µ 1 > µ 2 s 1 = 2.42 s 2 = 4.77 Stat → T statistics → Two sample → With summary Null: prop. diff.= Alternative Sample 1: Mean Std. Dev. Size Sample 2: Mean Std. Dev. Size ● Hypothesis Test P-value = ≠ (No pooled variance) Using StatCrunch (Be sure to not use pooled variance) Initial Conclusion: Since P-value < α (0.1), reject H 0 Final Conclusion: We accept the claim that the average age of students in the first class is greater than that in the second.

19 19 df = min(n 1 –1, n 2 –1) = min(92, 66) = 66 t α/2 = t 0.1/2 = t 0.05 = x 1 - x 2 = 21.2 – 19.8 = 1.4 (x 1 - x 2 ) + E = = (x 1 - x 2 ) – E = 1.4 – = CI = (0.34, 2.46) n 1 = 93 n 2 = 67α = 0.1 x 1 = 21.2 x 2 = 19.8 s 1 = 2.42 s 2 = 4.77 Example 3b mp (90% Confidence Interval)

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


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