Chapter 9. Inferences from Two Samples. Objective. Notation. Section 9.2. Definition. Notation. q = 1 p. Inferences About Two Proportions

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1 Chapter 9 Inferences from Two Samples 9. Inferences About Two Proportions 9.3 Inferences About Two s (Independent) 9.4 Inferences About Two s (Matched Pairs) 9.5 Comparing Variation in Two Samples Objective Compare the parameters of two populations using two samples from each population. Use Confidence Intervals and Hypothesis Tests For the first population use index 1 For the second population use index 9. Compare p 1, p 9.3 Compare µ 1, µ (Independent) 9.4 Compare µ 1, µ (Matched Pairs) 9.5 Compare σ 1, σ 1 Objective Section 9. Inferences About Two Proportions Compare the proportions of two populations using two samples from each population. Hypothesis Tests and Confidence Intervals of two proportions use the z-distribution 1 First Population p 1 x 1 p 1 Notation First population proportion First sample size Number of successes in first sample First sample proportion 3 4 Second Population p Notation Second population proportion Definition The pooled sample proportion p n x p Second sample size Number of successes in second sample Second sample proportion x 1 + x p = + n q = 1 p 5 6

2 Requirements (1) Have two independent random samples () For each sample: The number of successes is at least 5 The number of failures is at least 5 Both requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval Tests for Two Proportions The goal is to compare the two proportions H 0 : p 1 = p H 1 : p 1 p Two tailed H 0 : p 1 = p H 1 : p 1 < p Left tailed H 0 : p 1 = p H 1 : p 1 > p Right tailed Note: We only test the relation between p 1 and p (not the actual numerical values) 7 8 Finding the z = ( p ^ p ^ ) ( p p ) 1 1 pq + pq n Note: p 1 p =0 according to H 0 This equation is an altered form of the test statistic for a single proportion (see Ch. 8-3) 9 Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics) 10 Steps for Performing a Hypothesis Test on Two Proportions Write what we know State H 0 and H 1 Draw a diagram Calculate the sample and pooled proportions Find the Find the Critical Value(s) State the Initial Conclusion and Final Conclusion Note: Same process as in Chapter 8 11 Example 1 The table below lists results from a simple random sample of front-seat occupants involved in car crashes. Use a 0.05 significance level to test the claim that the fatality rate of occupants is lower for those in cars equipped with airbags. p 1 : Proportion of fatalities with airbags p : Proportion of fatalities with no airbags Claim p 1 < p What we know: x 1 = 41 x = 5 α = 0.05 = n = 9853 Claim: p 1 < p Note: Each sample has more than 5 successes and failures, thus fulfilling the requirements 1

3 Example 1 H 0 : p 1 = p H 1 : p 1 < p Sample Proportions Given: x 1 = 41 x = 5 α = 0.05 = n = 9853 Claim: p 1 < p Left-Tailed H 1 = Claim Diagram z = z α = Pooled Proportion z-dist. Example 1 H 0 : p 1 = p H 1 : p 1 < p Given: x 1 = 41 x = 5 α = 0.05 = n = 9853 Claim: p 1 < p Left-Tailed H 1 = Claim Diagram z = z α = Stat Proportions Two sample With summary Sample 1: Number of successes:. 41 Hypothesis Test Number of observations: Null: prop. diff.= Sample : Number of successes:. 5 Alternative Number of observations: < z-dist. P-value = 0.08 Critical Value () Initial Conclusion: Since z is in the critical region, reject H 0 Final Conclusion: We Accept the claim the fatality rate of occupants is lower for those who wear seatbelts 13 Initial Conclusion: Since P-value is less than α (with α = 0.05), reject H 0 Final Conclusion: We Accept the claim the fatality rate of occupants is lower for those who wear seatbelts 14 Confidence Interval Estimate We can observe how the two proportions relate by looking at the Confidence Interval Estimate of p 1 p CI = ( (p 1 p ) E, (p 1 p ) + E ) 3 Example 90% Confidence Interval Estimate of the difference x 1 = 41 x = 5 p 1 = = n = 9853 p = Where CI = ( , ) 15 Note: CI negative implies p 1 p is negative. This implies p 1 <p 16 Example 90% Confidence Interval Estimate of the difference Example 90% Confidence Interval Estimate of the difference x 1 = 41 x = 5 p 1 = = n = 9853 p = x 1 = 41 x = 5 = n = 9853 Stat Proportions Two sample With summary Sample 1: Number of successes:. 41 Confidence Interval Number of observations: Sample : Number of successes:. 5 Number of observations: 9853 Level 0.9 CI = ( , ) CI = ( , ) Note: CI negative implies p 1 p is negative. This implies p 1 <p 17 Note: CI negative implies p 1 p is negative. This implies p 1 <p 18

4 Interpreting Confidence Intervals If a confidence interval limits does not contain 0, it implies there is a significant difference between the two proportions (i.e. p 1 p ). Thus, we can interpret a relation between the two proportions from the confidence interval. In general: If p 1 = p then the CI should contain 0 If p 1 > p then the CI should be mostly positive If p 1 > p then the CI should be mostly negative 19 Example 3 Drug Clinical Trial Chantix is a drug used as an aid to stop smoking. The number of subjects experiencing insomnia for each of two treatment groups in a clinical trial of the drug Chantix are given below: Number in group Number experiencing insomnia Chantix Treatment (a) Use a 0.01 significance level to test the claim proportions of subjects experiencing insomnia is the same for both groups. (b) Find the 99% confidence level estimate of the difference of the two proportions. Does it support the result of the test? What we know: x 1 = 41 x = 5 α = 0.01 = 19 n = 9853 Claim: p 1 = p Note: Each sample has more than 5 successes and failures, thus fulfilling the requirements Placebo Example 3a H 0 : p 1 = p H 1 : p 1 p Sample Proportions Two-Tailed H 0 = Claim Given: x 1 = 19 x = 13 α = 0.01 = 19 n = 805 Claim: p 1 = p Diagram -z α/ = Pooled Proportion z-dist. z α/ =.576 z = Example 3a H 0 : p 1 = p H 1 : p 1 p Two-Tailed H 0 = Claim Stat Proportions Two sample With summary Sample 1: Number of successes:. Number of observations: Sample : Number of successes:. Number of observations: Given: x 1 = 19 x = 13 α = 0.01 = 19 n = 805 Claim: p 1 = p Diagram Hypothesis Test Null: prop. diff.= Alternative 0 z-dist. Critical Value () P-value < i.e. the P-value is very small Initial Conclusion: Since z is in the critical region, reject H 0 Initial Conclusion: Since the P-value is less than α (0.01), reject H 0 Final Conclusion: We Reject the claim the proportions of the subjects experiencing insomnia is the same in both groups. 1 Final Conclusion: We Reject the claim the proportions of the subjects experiencing insomnia is the same in both groups. Example 3b Use the same sample data in Example 3 to construct a 99% Confidence Interval Estimate of the difference Example 3b Use the same sample data in Example 3 to construct a 99% Confidence Interval Estimate of the difference x 1 = 19 x = 13 p 1 = = 19 n = 805 p = x 1 = 19 x = 13 = 19 n = 805 Stat Proportions Two sample With summary Sample 1: Number of successes:. Number of observations: Sample : Number of successes:. Number of observations: Confidence Interval Level 0.9 CI = (0.0500, 0.13 ) Note: CI does not contain 0 implies p 1 and p have significant difference. 3 CI = (0.0500, 0.13 ) Note: CI does not contain 0 implies p 1 and p have significant difference. 4

5 Section 9.3 Inferences About Two s (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 5 6 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) 5 First Population μ 1 σ 1 x 1 s 1 Notation First population mean First population standard deviation First sample size First sample mean First sample standard deviation 7 8 Second Population Notation Requirements (1) Have two independent random samples μ σ n x s Second population mean Second population standard deviation Second sample size Second sample mean Second sample standard deviation () σ 1 and σ are unknown and no assumption is made about their equality (3) Either or both the following holds: Both sample sizes are large ( >30, n >30) or Both populations have normal distributions All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval 9 30

6 Tests for Two Independent s Finding the The goal is to compare the two s H 0 : μ 1 = μ H 1 : μ 1 μ H 0 : μ 1 = μ H 1 : μ 1 < μ H 0 : μ 1 = μ H 1 : μ 1 > μ t x x 1 1 s 1 s n Two tailed Left tailed Right tailed Note: 1 =0 according to H 0 Degrees of freedom: df = smaller of 1 and n 1. Note: We only test the relation between μ 1 and μ (not the actual numerical values) 31 This equation is an altered form of the test statistic for a single mean when σ unknown (see Ch. 8-5) 3 Steps for Performing a Hypothesis Test on Two Independent s Write what we know Degrees of freedom df = min( 1, n 1) 6 State H 0 and H 1 Draw a diagram Find the Find the Degrees of Freedom Find the Critical Value(s) State the Initial Conclusion and Final Conclusion Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics) 33 Note: Same process as in Chapter 8 34 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 = 186 n = 10 α = 0.05 x 1 = x = Claim: μ 1 = μ H 0 : µ 1 = µ H 1 : µ 1 µ s 1 = s = Two-Tailed H 0 = Claim t = t α/ = t-dist. df = 185 t α/ = 1.97 Degrees of Freedom df = min( 1, n 1) = min(185, 09) = 185 Critical Value t α/ = t 0.05 = 1.97 () 35 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. 36

7 Example 1 = 186 n = 10 α = 0.05 x 1 = x = Claim: μ 1 = μ H 0 : µ 1 = µ H 1 : µ 1 µ Two-Tailed H 0 = Claim (Be sure to not use pooled variance) s 1 = s = Stat T statistics Two sample With summary Sample 1: Sample : Hypothesis Test Null: prop. diff.= Alternative 0 10 (No pooled variance) P-value = Confidence Interval Estimate We can observe how the two proportions relate by looking at the Confidence Interval Estimate of μ 1 μ CI = ( (x 1 x ) E, (x 1 x ) + E ) Where 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. 37 df = min( 1, n 1) 38 Example 95% Confidence Interval Estimate of the difference between the two population proportions (µ 1 µ ) = 186 n = 10 df = min(n df = min( 1, n 1) = min(185, 10) = , n 1) = min(185, 10) = 185 x 1 = x = t t α/ t 0.1/ = t 0.05 = α/ = t 0.05/ = t 0.05 = s 1 = s = x x 1 - x = = x = = Example 95% Confidence Interval Estimate of the difference between the two population proportions (µ 1 µ ) = 186 n = 10 x 1 = x = s 1 = s = Stat T statistics Two sample With summary Sample 1: Sample : Confidence Interval Level: 0.95 (No pooled variance) (x 1 - x ) + E = = (x 1 - x ) E = = CI = (-14.7, ) 39 CI = (-137.4, ) Note: slightly different because of rounding errors 40 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: = 93 n = 67 x 1 = 1. x = 19.8 s 1 =.4 s = 4.77 (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. 41 Example 3a H 0 : µ 1 = µ H 1 : µ 1 > µ t = x 1 x s1 n1 +s n Critical Value t α/ = t 0.05 = Right-Tailed H 1 = Claim = 93 n = 67 α = 0.1 = =.07 (.4) 93 +(4.77) 67 Degrees of Freedom df = min( 1, n 1) = min(9, 66) = 66 x 1 = 1. x = 19.8 Claim: µ 1 > µ s 1 =.4 s = 4.77 () Initial Conclusion: Since t is in the critical region, reject H 0 t α/ = t-dist. df = 66 t = 7.60 Final Conclusion: We accept the claim that the average age of students in the first class is greater than that in the second. 4

8 Example 3a = 93 n = 67 α = 0.1 x 1 = 1. x = 19.8 Claim: µ 1 > µ H 0 : µ 1 = µ H 1 : µ 1 > µ Right-Tailed H 1 = Claim (Be sure to not use pooled variance) s 1 =.4 s = 4.77 Stat T statistics Two sample With summary Sample 1: Sample : 1. Hypothesis Test.4 Null: prop. diff.= 0 93 Alternative (No pooled variance) P-value = Example 3b (90% Confidence Interval) df = min( 1, n 1) = min(9, 66) = 66 t α/ = t 0.1/ = t 0.05 = x 1 - x = = 1.4 s E = t 1 α + s 1 = mp (x 1 - x ) + E = =.458 (x 1 - x ) E = = 0.34 = 93 n = 67 α = 0.1 x 1 = 1. x = 19.8 s 1 =.4 s = = Initial Conclusion: Since P-value < α (0.1), reject H 0 CI = (0.34,.46) 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 (90% Confidence Interval) (Be sure to not use pooled variance) = 93 n = 67 α = 0.1 x 1 = 1. x = 19.8 s 1 =.4 s = 4.77 Stat T statistics Two sample With summary Sample 1: Sample : 1. Hypothesis Test.4 Null: prop. diff.= 0 93 Alternative (No pooled variance) 8 CI = (0.35,.45) 45