Chapter 9 Slide Ifereces from Two Samples 9- Overview 9- Ifereces about Two Proportios 9- Ifereces about Two Meas: Idepedet Samples 9-4 Ifereces about Matched Pairs 9-5 Comparig Variatio i Two Samples Overview Slide There are may importat ad meaigful situatios i which it becomes ecessary to compare two sets of sample data. Ifereces about Two Proportios Assumptios Slide. We have proportios from two idepedet simple radom samples.. For both samples, the coditios p 5 ad q 5 are satisfied. Notatio for Two Proportios q = p The correspodig meaigs are attached to p,, x, p. ad q, which come from populatio. Slide 4 For populatio, we let: p = populatio proportio = size of the sample x = umber of successes i the sample p = x (the sample proportio) The pooled estimate of p ad p is deoted by p. q = p Pooled Estimate of p ad p p = x x Slide 5 Test Statistic for Two Proportios For H 0 : p = p, H 0 : p p, H 0 : p p H : p p, H : p < p, H : p > p z = ( p p ) ( p p ) pq pq Slide 6
Test Statistic for Two Proportios For H 0 : p = p, H 0 : p p, H 0 : p p Slide 7 Example: For the sample data listed i Table 8-, use a 0.05 sigificace level to test Slide 8 the claim that the proportio of black drivers stopped by the police is greater tha the proportio of white drivers H : p p, H : p < p, H : p > p where p p = 0 (assumed i the ull hypothesis) p x = ad p x = p = x x ad q = p Example: For the sample data listed i Table 8-, use a 0.05 sigificace level to test Slide 9 the claim that the proportio of black drivers stopped by the police is greater tha the proportio of white drivers = 00 x = 4 p = x = 4 = 0.0 00 = 400 x = 47 p = x = 47 = 0.05 400 H 0 : p = p, H : p > p p = x x = 4 47 = 0.06875 00400 q = 0.06875 = 0.895. Example: For the sample data listed i Table 8-, use a 0.05 sigificace level to test Slide 0 the claim that the proportio of black drivers stopped by the police is greater tha the proportio of white drivers = 00 x = 4 p = x = 4 = 0.0 00 = 400 x = 47 p = x = 47 = 0.05 400 z = (0.0 0.05) 0 (0.06875)(0.895) (0.06875)(0.895) 00 400 z = 0.64 Example: For the sample data listed i Table 8-, use a 0.05 sigificace level to test Slide the claim that the proportio of black drivers stopped by the police is greater tha the proportio of white drivers = 00 x = 4 p = x = 4 = 0.0 00 = 400 x = 47 p = x = 47 = 0.05 400 z = 0.64 This is a right-tailed test, so the P- value is the area to the right of the test statistic z = 0.64. The P-value is 0.6. Because the P-value of 0.6 is greater tha the sigificace level of α = 0.05, we fail to reject the ull hypothesis. Example: For the sample data listed i Table 8-, use a 0.05 sigificace level to test Slide the claim that the proportio of black drivers stopped by the police is greater tha the proportio of white drivers = 00 x = 4 p = x = 4 = 0.0 00 = 400 x = 47 p = x = 47 = 0.05 400 z = 0.64 Because we fail to reject the ull hypothesis, we coclude that there is ot sufficiet evidece to support the claim that the proportio of black drivers stopped by police is greater tha that for white drivers. This does ot mea that racial profilig has bee disproved. The evidece might be strog eough with more data.
Example: For the sample data listed i Table 8-, use a 0.05 sigificace level to test Slide the claim that the proportio of black drivers stopped by the police is greater tha the proportio of white drivers = 00 x = 4 p = x = 4 = 0.0 00 = 400 x = 47 p = x = 47 = 0.05 400 Usig TI Calculator: Slide 4 4 Compare these results with the example preseted i the last few slides. Cofidece Iterval Estimate of p - p Slide 5 Example: For the sample data listed i Table 8-, fid a 90% cofidece iterval estimate of the differece betwee the two populatio proportios. Slide 6 ( p p ) E < ( p p ) < ( p p ) E where E = z α/ p q p q = 00 x = 4 p = x = 4 = 0.0 00 = 400 x = 47 p = x = 47 = 0.05 400 E = z α/ E =.645 E = 0.400 p q p q (.)(.88)(0.05)(0.895) 00 400 Example: For the sample data listed i Table 8-, use a 0.05 sigificace level to test Slide 7 the claim that the proportio of black drivers stopped by the police is greater tha the proportio of white drivers Usig TI Calculator: Slide 8 = 00 (0.0 0.05) 0.040 < ( p p ) < (0.0 0.05) 0.040 0.05 < ( p x = 4 p ) < 0.055 p = x = 4 = 0.0 00 = 400 x = 47 p = x = 47 = 0.05 400 4 Compare these results with the example preseted i the last few slides.
Defiitios Slide 9 Assumptios Slide 0 Two Samples: Idepedet The sample values selected from oe populatio are ot related or somehow paired with the sample values selected from the other populatio. If the values i oe sample are related to the values i the other sample, the samples are depedet. Such samples are ofte referred to as matched pairs or paired samples.. The two samples are idepedet.. Both samples are simple radom samples.. Either or both of these coditios are satisfied: The two sample sizes are both large (with > 0 ad > 0) or both samples come from populatios havig ormal distributios. Hypothesis Tests Slide Hypothesis Tests Slide Test Statistic for Two Meas: Test Statistic for Two Meas: t = (x x ) (µ µ ). s s 4 Degrees of freedom: I this book we use this estimate: df = smaller of ad. P-value: Critical values: Refer to Table A-. Use the procedure summarized i Figure 7-6. Refer to Table A-. McGwire Versus Bods Slide McGwire Versus Bods Slide 4 Data Set 0 i Appedix B icludes the distaces of the home rus hit i record-settig seasos by Mark McGwire ad Barry Bods. Sample statistics are show. Use a 0.05 sigificace level to test the claim that the distaces come from populatios with differet meas. McGwire Bods 70 7 x 48.5 40.7 s 45.5 0.6 Claim: μ μ H o : μ = μ H : μ μ α = 0.05 = 69 = 7 df = 69 t.05 =.994
McGwire Versus Bods McGwire Versus Bods Slide 5 Slide 6 Test Statistic for Two Meas: Test Statistic for Two Meas: t = (x x ) (µ µ ). s s t = (48.5 40.7) 0 45.5 0.6 70 7 =.7 Claim: μ μ H o : μ = μ H : μ μ α = 0.05 McGwire Versus Bods Slide 7 5 Claim: μ μ H o : μ = μ H : μ μ α = 0.05 McGwire Versus Bods Slide 8 There is sigificat evidece to support the claim that there is a differece betwee the mea home ru distaces of Mark McGwire ad Barry Bods. Reject Null Figure 8- Figure 8- Usig TI Calculator: Slide 9 Cofidece Itervals Slide 0 (x x ) E < (µ µ ) < (x x ) E 4 s s where E = tα/ Our textbook recommeds to use No Pooled optio at this time.
McGwire Versus Bods McGwire Versus Bods Slide Slide Usig the sample data give i the precedig example, costruct a 95%cofidece iterval estimate of the differece betwee the mea home ru distaces of Mark McGwire ad Barry Bods. E = t s s α/ E =.994 E =.0 45.5 0.6 70 7 Usig the sample data give i the precedig example, costruct a 95%cofidece iterval estimate of the differece betwee the mea home ru distaces of Mark McGwire ad Barry Bods. (48.5 40.7).0 < (μ μ ) < (48.5 40.7).0.8 < (μ μ ) < 7.8 We are 95% cofidet that the limits of.8 ft ad 7.8 ft actually do cotai the differece betwee the two populatio meas. Usig TI Calculator: Slide Assumptios Slide 4. The sample data cosist of matched pairs. 4 6. The samples are simple radom samples.. Either or both of these coditios is satisfied: The umber of matched pairs of sample data is ( > 0) or the pairs of values have differeces that are from a populatio havig a distributio that is approximately ormal. Slide 5 µ d = mea value of the differeces d for the populatio of paired data d Notatio for Matched Pairs = mea value of the differeces d for the paired sample data (equal to the mea of the x y values) Test Statistic for Matched Pairs of Sample Data t = d µ d s d Slide 6 s d = stadard deviatio of the differeces d for the paired sample data = umber of pairs of data. where degrees of freedom =
P-values ad Critical Values Slide 7 Cofidece Itervals Slide 8 Use Table A- (t-distributio). d E < µ d < d E where E = t α/ s d degrees of freedom = Slide 9 Slide 40 d =., s = 0.7, = 5 t α/ =.776 (foud from Table A- with 4 degrees of freedom ad 0.05 i two tails) 7 H 0 : μ d = 0 H : μ d 0 t = d µ d =. 0 =.759 s d 0.7 5 Because the test statistic does ot fall i the critical regio, we fail to reject the ull hypothesis. Slide 4 Slide 4 H 0 : μ d = 0 H : μ d 0 t = d µ d =. 0 =.759 s d 0.7 5 The sample data i Table 8- do ot provide sufficiet evidece to support the claim that actual ad five-day forecast low temperatures are differet.
Slide 4 Slide 44 Usig the same sample matched pairs i Table 8-, costruct a 95% cofidece iterval estimate of μ d, which is the mea of the differeces betwee actual low temperatures ad five-day forecasts. E = t α/ s d 0.7 E = (.776)( 5 ) =. Slide 45 Slide 46 d E < μ d < d E.. < μ d <.. 6.5 < μ d < 0. 8 I the log ru, 95% of such samples will lead to cofidece itervals that actually do cotai the true populatio mea of the differeces. Usig TI Calculator: Slide 47 Usig TI Calculator: Slide 48 Eter actual low ito L, predicted values ito L, ad the differece ito L as explaied i class. 5 6 4 7 8 Compare these results with the example preseted i the last few slides.
Measures of Variatio Slide 49 Assumptios Slide 50 s = stadard deviatio of sample σ = stadard deviatio of populatio s = variace of sample σ = variace of populatio. The two populatios are idepedet of each other.. The two populatios are each ormally distributed. s σ Notatio for Hypothesis Tests with Two Variaces = larger of the two sample variaces = size of the sample with the larger variace Slide 5 = variace of the populatio from which the sample with the larger variace was draw The symbols s,, ad σ are used for the other sample ad populatio. 9 Test Statistic for Hypothesis Tests with Two Variaces s F = s Critical Values: Usig Table A-5, we obtai critical F values that are determied by the followig three values: Slide 5. The sigificace level α.. Numerator degrees of freedom (df ) =. Deomiator degrees of freedom (df ) = Slide 5 All oe-tailed tests will be right-tailed. All two-tailed tests will eed oly the critical value to the right. Whe degrees of freedom is ot listed exactly, use the critical values o either side as a iterval. Use iterpolatio oly if the test statistic falls withi the iterval. Slide 54 If the two populatios do have equal s variaces, the F= s will be close to because s ad s are close i value. If the two populatios have radically differet variaces, the F will be a large umber. Remember, the larger sample variace will be s.
Slide 55 Coke Versus Pepsi Slide 56 Cosequetly, a value of F ear will be evidece i favor of the coclusio that σ = σ. But a large value of F will be evidece agaist the coclusio of equality of the populatio variaces. Data Set 7 i Appedix B icludes the weights (i pouds) of samples of regular Coke ad regular Pepsi. Sample statistics are show. Use the 0.05 sigificace level to test the claim that the weights of regular Coke ad the weights of regular Pepsi have the same stadard deviatio. Regular Coke Regular Pepsi 6 6 x 0.868 0.840 s 0.007507 0.00570 Coke Versus Pepsi Slide 57 Coke Versus Pepsi Slide 58 Claim: σ = σ Claim: σ = σ H o : σ = σ H : σ σ α = 0.05 0 H o : σ = σ H : σ σ α = 0.05 Value of F = s s 0.007507 = 0.00570 =.79 There is ot sufficiet evidece to warrat rejectio of the claim that the two variaces are equal. Usig TI Calculator: Slide 59 Fidig Lower Critical F Values Slide 60 4Draw ) Use F R idicates the critical value for right tail ad F L idicates the critical value for the left tail. ) Iterchage the degrees of freedom. ) F L is the reciprocal of the F value foud i the table. For example: =7, =0, α=0.05 Aswer: F L =0.80, F R =4.97