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Chapter 10 Statistical Inference About Means and Proportions With Two Populations Slide 1 Learning objectives 1. Understand inferences About the Difference Between Two Population Means: σ 1 and σ 2 Known 2. Understand Inferences About the Difference Between Two Population Means: σ 1 and σ 2 Unknown 3. Understand Inferences About the Difference Between Two Population Means: Matched Samples 4. Understand Inferences About the Difference Between Two Population Proportions Slide 2 1 Overview Tests for two population Two means (Point estimator = x1 − x2 ) Two proportions (Point estimator = p1 − p2) σ1 and σ2 are known Standard error will be different for each case. σ1 and σ2 are unknown Matched (or paired) sample Slide 3 Comparison between one population case and two population case. nProcedures of getting confidence interval and of testing for two population case is very similar as those of one population case nConfidence interval: Point estimator ± Significance coefficient * Standard Error nTest Statistics Point estimator - Hypothesized value of parameter Standard error nPoint estimators and standard errors of two population case are different from those of one population case. Slide 4 2 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Estimating the Difference Between Two Population Means Let µ1 equal the mean of population 1 and µ2 equal the mean of population 2. n The difference between the two population means is µ1 - µ2. n Let x1 equal the mean of sample 1 and x2 equal the mean of sample 2. n The point estimator of the difference between the means of the populations 1 and 2 is x1 − x2. Slide 5 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Sampling Distribution of x1 − x2 : σ known Expected Value E ( x1 − x 2 ) = µ 1 − µ 2 n Standard Deviation (Standard Error) σ x1 − x2 = σ12 σ 22 + n1 n2 where: σ1 = standard deviation of population 1 σ2 = standard deviation of population 2 n1 = sample size from population 1 n2 = sample size from population 2 Slide 6 3 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Interval Estimation of µ1 - µ2: σ 1 and σ 2 Known Interval Estimate x1 − x2 ± zα / 2 σ12 σ 22 + n1 n2 where: 1 - α is the confidence coefficient Slide 7 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Interval Estimation of µ1 - µ2: σ 1 and σ 2 Known n Example: Par, Inc. Par, Inc. is a manufacturer of golf equipment and has developed a new golf ball that has been designed to provide “extra distance.” In a test of driving distance using a mechanical driving device, a sample of Par golf balls was compared with a sample of golf balls made by Rap, Ltd., a competitor. The sample statistics appear on the next slide. Slide 8 4 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Interval Estimation of µ1 - µ2: σ 1 and σ 2 Known Example: Par, Inc. Sample Size Sample Mean Sample #1 Par, Inc. 120 balls 275 yards Sample #2 Rap, Ltd. 80 balls 258 yards Based on data from previous driving distance tests, the two population standard deviations are known with σ 1 = 15 yards and σ 2 = 20 yards. Slide 9 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Interval Estimation of µ1 - µ2: σ 1 and σ 2 Known Let us develop a 95% confidence interval estimate of the difference between the mean driving distances of the two brands of golf ball. Point estimate of µ1 − µ2 = x1 − x2 = 275 − 258 = 17 yards where: µ1 = mean distance for the population of Par, Inc. golf balls µ2 = mean distance for the population of Rap, Ltd. golf balls Slide 10 5 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Interval Estimation of µ1 - µ2: σ 1 and σ 2 Known x1 − x2 ± zα / 2 σ12 σ 22 (15) 2 ( 20) 2 + = 17 ± 1. 96 + n1 n2 120 80 17 + 5.14 or 11.86 yards to 22.14 yards We are 95% confident that the difference between the mean driving distances of Par, Inc. balls and Rap, Ltd. balls is 11.86 to 22.14 yards. Slide 11 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Known n Hypotheses H0 : µ1 − µ2 ≥ D0 H0 : µ1 − µ2 ≤ D0 H0 : µ1 − µ2 = D0 Ha : µ1 − µ2 < D0 Ha : µ1 − µ2 > D0 Ha : µ1 − µ2 ≠ D0 Left-tailed Right-tailed Two-tailed n Test Statistic z= ( x1 − x2 ) − D0 σ 12 σ 22 + n1 n2 Slide 12 6 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Known Example: Par, Inc. Can we conclude, using α = .01, that the mean driving distance of Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls? Slide 13 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Known n p –Value and Critical Value Approaches 1. Develop the hypotheses. H0 : µ 1 - µ 2 < 0 Ha : µ 1 - µ 2 > 0 where: µ1 = mean distance for the population of Par, Inc. golf balls µ2 = mean distance for the population of Rap, Ltd. golf balls 2. Specify the level of significance. α = .01 Slide 14 7 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Known n p –Value and Critical Value Approaches 3. Compute the value of the test statistic. z= z= ( x 1 − x 2 ) − D0 σ 12 σ 22 + n1 n2 (235 − 218) − 0 2 (15) (20 ) + 120 80 2 = 17 = 6.49 2.62 Slide 15 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Known n p –Value Approach 4. Compute the p–value. For z = 6.49, the p –value < .0001. 5. Determine whether to reject H0. Because p–value < α = .01, we reject H0. At the .01 level of significance, the sample evidence indicates the mean driving distance of Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls. Slide 16 8 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Known n Critical Value Approach 4. Determine the critical value and rejection rule. For α = .01, z.01 = 2.33 Reject H0 if z > 2.33 5. Determine whether to reject H0. Because z = 6.49 > 2.33, we reject H0. The sample evidence indicates the mean driving distance of Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls. Slide 17 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n n In-class exercise #2 (p.400) (confidence interval) #3 (p.400) (hypothesis testing) Slide 18 9 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Interval Estimation of µ1 - µ2: σ 1 and σ 2 Unknown When σ 1 and σ 2 are unknown, we will: • use the sample standard deviations s1 and s2 as estimates of σ 1 and σ 2 , and • replace zα/2 with tα/2. Slide 19 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Interval Estimation of µ1 - µ2: σ 1 and σ 2 Unknown Interval Estimate x1 − x2 ± tα / 2 s12 s22 + n1 n2 Where the degrees of freedom for tα/2 are: 2 s12 s22 + n1 n2 df = 2 2 1 s12 1 s22 + n1 −1 n1 n2 −1 n2 Slide 20 10 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Difference Between Two Population Means: σ 1 and σ 2 Unknown Example: Specific Motors Specific Motors of Detroit has developed a new automobile known as the M car. 24 M cars and 28 J cars (from Japan) were road tested to compare miles-per-gallon (mpg) performance. The sample statistics are shown on the next slide. Slide 21 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Difference Between Two Population Means: σ 1 and σ 2 Unknown Example: Specific Motors Sample #1 M Cars 24 cars 29.8 mpg 2.56 mpg Sample #2 J Cars 28 cars 27.3 mpg 1.81 mpg Sample Size Sample Mean Sample Std. Dev. Slide 22 11 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Difference Between Two Population Means: σ 1 and σ 2 Unknown Example: Specific Motors Let us develop a 90% confidence interval estimate of the difference between the mpg performances of the two models of automobile. Slide 23 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Point Estimate of µ 1 − µ 2 Point estimate of µ1 − µ2 = x1 − x2 = 29.8 - 27.3 = 2.5 mpg where: µ1 = mean miles-per-gallon for the population of M cars µ2 = mean miles-per-gallon for the population of J cars Slide 24 12 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Interval Estimation of µ 1 − µ 2: σ 1 and σ 2 Unknown The degrees of freedom for tα/2 are: 2 (2.56)2 (1.81)2 + 24 28 df = = 24.07 = 24 2 2 2 2 1 (2.56) 1 (1.81) + 24 −1 24 28 −1 28 With α/2 = .05 and df = 24, tα/2 = 1.711 Slide 25 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Interval Estimation of µ 1 − µ 2: σ 1 and σ 2 Unknown x1 − x2 ± tα / 2 s12 s22 (2.56)2 (1.81)2 + = 29.8 − 27.3 ± 1.711 + n1 n2 24 28 2.5 + 1.069 or 1.431 to 3.569 mpg We are 90% confident that the difference between the miles-per-gallon performances of M cars and J cars is 1.431 to 3.569 mpg. Slide 26 13 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Unknown Hypotheses n H0 : µ1 − µ2 ≥ D0 H0 : µ1 − µ2 ≤ D0 H0 : µ1 − µ2 = D0 Ha : µ1 − µ2 < D0 Ha : µ1 − µ2 > D0 Ha : µ1 − µ2 ≠ D0 Left-tailed Right-tailed Two-tailed Test Statistic n t= ( x1 − x2 ) − D 0 s12 s 22 + n1 n2 Slide 27 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Unknown Example: Specific Motors Can we conclude, using a .05 level of significance, that the miles-per-gallon (mpg) performance of M cars is greater than the miles-pergallon performance of J cars? Slide 28 14 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Unknown n p –Value and Critical Value Approaches 1. Develop the hypotheses. H0 : µ 1 - µ 2 < 0 Ha : µ 1 - µ 2 > 0 where: µ1 = mean mpg for the population of M cars µ2 = mean mpg for the population of J cars Slide 29 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Unknown n p –Value and Critical Value Approaches 2. Specify the level of significance. α = .05 3. Compute the value of the test statistic. t= ( x1 − x2 ) − D0 2 1 2 2 s s + n1 n2 = (29.8 − 27.3) − 0 (2.56) 2 (1.81) 2 + 24 28 = 4.003 Slide 30 15 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Unknown n p –Value Approach 4. Compute the p –value. The degrees of freedom for tα are: 2 df = (2.56)2 (1.81)2 + 28 24 2 2 1 (2.56)2 1 (1.81)2 + 24 − 1 24 28 − 1 28 = 24.07 = 24 Because t = 4.003 > t.005 = 2.797, the p–value < .005. Slide 31 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Unknown n p –Value Approach 5. Determine whether to reject H0. Because p–value < α = .05, we reject H0. We are at least 95% confident that the miles-pergallon (mpg) performance of M cars is greater than the miles-per-gallon performance of J cars?. Slide 32 16 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests About µ 1 − µ 2: σ 1 and σ 2 Unknown n Critical Value Approach 4. Determine the critical value and rejection rule. For α = .05 and df = 24, t.05 = 1.711 Reject H0 if t > 1.711 5. Determine whether to reject H0. Because 4.003 > 1.711, we reject H0. We are at least 95% confident that the miles-pergallon (mpg) performance of M cars is greater than the miles-per-gallon performance of J cars?. Slide 33 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n n In-class exercise #9 (p.406) (confidence interval) #10 (p.407) (hypothesis testing) Slide 34 17 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Inferences About the Difference Between Two Population Means: Matched Samples n With a matched-sample design each sampled item provides a pair of data values. n This design often leads to a smaller sampling error than the independent-sample design because variation between sampled items is eliminated as a source of sampling error. Slide 35 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Inferences About the Difference Between Two Population Means: Matched Samples Example: Express Deliveries A Chicago-based firm has documents that must be quickly distributed to district offices throughout the U.S. The firm must decide between two delivery services, UPX (United Parcel Express) and INTEX (International Express), to transport its documents. Slide 36 18 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Inferences About the Difference Between Two Population Means: Matched Samples Example: Express Deliveries In testing the delivery times of the two services, the firm sent two reports to a random sample of its district offices with one report carried by UPX and the other report carried by INTEX. Do the data on the next slide indicate a difference in mean delivery times for the two services? Use a .05 level of significance. Slide 37 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Inferences About the Difference Between Two Population Means: Matched Samples Delivery Time (Hours) District Office UPX INTEX Difference Seattle Los Angeles Boston Cleveland New York Houston Atlanta St. Louis Milwaukee Denver 32 30 19 16 15 18 14 10 7 16 25 24 15 15 13 15 15 8 9 11 7 6 4 1 2 3 -1 2 -2 5 Slide 38 19 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Inferences About the Difference Between Two Population Means: Matched Samples n p –Value and Critical Value Approaches 1. Develop the hypotheses. H0 : µ d = 0 Ha : µ d ≠ 0 Let µd = the mean of the difference values for the two delivery services for the population of district offices Slide 39 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Inferences About the Difference Between Two Population Means: Matched Samples n p –Value and Critical Value Approaches 2. Specify the level of significance. α = .05 3. Compute the value of the test statistic. d = sd = t= ∑ di ( 7 + 6 +... +5) = = 2. 7 n 10 2 76.1 ∑ ( di − d ) = = 2. 9 n −1 9 d − µd 2.7 − 0 = = 2.94 sd n 2.9 10 Slide 40 20 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Inferences About the Difference Between Two Population Means: Matched Samples n p –Value Approach 4. Compute the p –value. For t = 2.94 and df = 9, the p–value is between .02 and .01. (This is a two-tailed test, so we double the upper-tail areas of .01 and .005.) 5. Determine whether to reject H0. Because p–value < α = .05, we reject H0. We are at least 95% confident that there is a difference in mean delivery times for the two services? Slide 41 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Inferences About the Difference Between Two Population Means: Matched Samples n Critical Value Approach 4. Determine the critical value and rejection rule. For α = .05 and df = 9, t.025 = 2.262. Reject H0 if t > 2.262 5. Determine whether to reject H0. Because t = 2.94 > 2.262, we reject H0. We are at least 95% confident that there is a difference in mean delivery times for the two services? Slide 42 21 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n n In-class exercise #21 (p.414) (hypothesis testing) #22 (p.414) (confidence interval) Slide 43 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Sampling Distribution of p1 − p2 Expected Value E ( p1 − p2 ) = p1 − p2 n Standard Deviation (Standard Error) σ p1 − p2 = p1 (1 − p1 ) p2 (1 − p2 ) + n1 n2 where: n1 = size of sample taken from population 1 n2 = size of sample taken from population 2 Slide 44 22 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Sampling Distribution of p1 − p2 If the sample sizes are large, the sampling distribution of p1 − p2 can be approximated by a normal probability distribution. The sample sizes are sufficiently large if all of these conditions are met: n1p1 > 5 n1(1 - p1) > 5 n2p2 > 5 n2(1 - p2) > 5 Slide 45 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Sampling Distribution of p1 − p2 σ p1 − p2 = p1 (1 − p1 ) p2 (1 − p2 ) + n1 n2 p1 − p2 p1 – p2 Slide 46 23 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Interval Estimation of p1 - p2 Interval Estimate p1 − p2 ± zα / 2 p1 (1 − p1 ) p2 (1 − p2 ) + n1 n2 Slide 47 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Interval Estimation of p1 - p2 Example: Market Research Associates Market Research Associates is conducting research to evaluate the effectiveness of a client’s new advertising campaign. Before the new campaign began, a telephone survey of 150 households in the test market area showed 60 households “aware” of the client’s product. The new campaign has been initiated with TV and newspaper advertisements running for three weeks. Slide 48 24 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Interval Estimation of p1 - p2 Example: Market Research Associates n A survey conducted immediately after the new campaign showed 120 of 250 households “aware” of the client’s product. Does the data support the position that the advertising campaign has provided an increased awareness of the client’s product? Slide 49 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Point Estimator of the Difference Between Two Population Proportions p1 = proportion of the population of households “aware” of the product after the new campaign p2 = proportion of the population of households “aware” of the product before the new campaign p1 = sample proportion of households “aware” of the product after the new campaign p2 = sample proportion of households “aware” of the product before the new campaign p1 − p2 = 120 60 − = .48 − .40 = .08 250 150 Slide 50 25 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Interval Estimation of p1 - p2 For α = .05, z.025 = 1.96: .48 − .40 ± 1.96 .48(.52) .40(.60) + 250 150 .08 + 1.96(.0510) .08 + .10 Hence, the 95% confidence interval for the difference in before and after awareness of the product is -.02 to +.18. Slide 51 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n Hypothesis Tests about p1 - p2 Hypotheses We focus on tests involving no difference between the two population proportions (i.e. p1 = p2) H0 : p1 − p2 ≥ 0 Ha : p1 − p2 < 0 Left-tailed H H00:: H Ha:: a pp1 − - p2 < 0 1 p2 ≤ 0 pp1 − - pp2 >> 00 1 2 Right-tailed H0 : p1 − p2 = 0 Ha : p1 − p2 ≠ 0 Two-tailed Slide 52 26 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests about p1 - p2 n Pooled Estimate of Standard Error of p1 − p2 1 1 σ p1− p2 = p(1− p) + n1 n2 where: p= n1 p1 + n2 p2 n1 + n2 Slide 53 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests about p1 - p2 n Test Statistic z= ( p1 − p 2 ) 1 1 p (1 − p ) + n1 n2 Slide 54 27 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests about p1 - p2 Example: Market Research Associates Can we conclude, using a .05 level of significance, that the proportion of households aware of the client’s product increased after the new advertising campaign? n Slide 55 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests about p1 - p2 n p -Value and Critical Value Approaches 1. Develop the hypotheses. H0: p1 - p2 < 0 Ha: p1 - p2 > 0 p1 = proportion of the population of households “aware” of the product after the new campaign p2 = proportion of the population of households “aware” of the product before the new campaign Slide 56 28 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests about p1 - p2 n p -Value and Critical Value Approaches 2. Specify the level of significance. α = .05 3. Compute the value of the test statistic. p= 250(. 48) + 150(. 40) 180 = = . 45 250 + 150 400 s pp1 −−pp2 = . 45(. 55)( 1 + 1 ) =. 0514 250 150 1 2 z= (.48 − .40) − 0 .08 = = 1.56 .0514 .0514 Slide 57 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests about p1 - p2 n p –Value Approach 4. Compute the p –value. For z = 1.56, the p–value = .0594 5. Determine whether to reject H0. Because p–value > α = .05, we cannot reject H0. We cannot conclude that the proportion of households aware of the client’s product increased after the new campaign. Slide 58 29 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions Hypothesis Tests about p1 - p2 n Critical Value Approach 4. Determine the critical value and rejection rule. For α = .05, z.05 = 1.645 Reject H0 if z > 1.645 5. Determine whether to reject H0. Because 1.56 < 1.645, we cannot reject H0. We cannot conclude that the proportion of households aware of the client’s product increased after the new campaign. Slide 59 •L.O. •Test for means −σ known −σ unknown −Matched sample •Test for proportions n n In-class exercise #30 (p.420) (confidence interval) #31 (p.421) (hypothesis testing) Slide 60 30 End of Chapter 10 Slide 61 Formulas for two population case Confidence Interval Means: σ known Means: σ unknown Means: matched Proportion Test Statistics x1 − x2 ± zα /2 σ12 σ22 + n1 n2 z= x1 − x2 ± tα / 2 s12 s22 + n1 n2 t= Other formula ( x1 − x2 ) − D0 σ 12 σ 22 + n1 n2 ( x1 − x 2 ) − D 0 2 1 2 2 s s + n1 n 2 2 df = s12 s22 + n1 n2 2 2 1 s12 1 s22 + n1 −1 n1 n2 −1 n2 Same as one population case p1 − p2 ±zα/2 p1(1− p1) p2(1− p2) + n1 n2 z= ( p1 − p2 ) 1 1 p (1 − p ) + n1 n 2 p= n1p1 +n2p2 n1 +n2 Slide 62 31