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Two Population Means Hypothesis Testing and Confidence Intervals For Differences in Proportions
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SITUATION: 2 Populations Population 1 Population 2 Men - Republican?Women - Republican?TV in Cuba - Good?TV in China - Good?HS >$100K?College > $100K?Trad. Course “A”?Internet Course “A”? Yes = 1 No = 0
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Notation
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Individual Responses Proportion of Responses Each Individual Response is: 1 = YES or 0 = NO –Bernoulli Distribution: Mean = p; Variance = pq (where q = (1-p)) The Average or Proportion of n responses (n large) by the Central Limit Theorem is: –Distributed approximately normal with mean = p –Variance = pq/n
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Differences in Proportions Proportion of “1’s” from Population 1 has: –a normal distribution (approximately) –true mean: p 1 (which we don’t know) –true variance p 1 q 1 /n 1 Proportion of “1’s” from Population 2 has: –a normal distribution (approximately) –true mean: p 2 (which we don’t know) –true variance p 2 q 2 /n 2
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Distribution of the Difference in Proportions True mean: p 1 - p 2 True variance p 1 q 1 /n 1 + p 2 q 2 /n 2 True standard deviation: SQRT(p 1 q 1 /n 1 + p 2 q 2 /n 2 ) –But since p 1 and p 2 are unknown, what should we use for the standard deviation in confidence intervals and hypothesis tests? For all confidence intervals and for all hypothesis tests except H0: p1- p2 = 0 For hypothesis tests of the form H0: p1- p2 = 0
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Hypothesis Tests and Confidence Intervals H 0 : p 1 - p 2 = v Where v ≠ 0 H 0 : p 1 - p 2 = 0 Z-Statistics for Difference in Proportions Confidence Interval
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Example Midas wants to compare customer satisfaction between NY and LA operations. Can it conclude: (1) A greater proportion in LA are satisfied? (2) Customer satisfaction in LA exceeds NY by > 2%? (3) Give a 95% confidence interval for difference in customer satisfaction. Results -- 350 out of 400 in LA were satisfied 160 out of 200 in NY were satisfied
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(1) Is a greater proportion in LA? H 0 : p 1 - p 2 = 0 H A : p 1 - p 2 > 0 Select α =.05 Reject H 0 (Accept H A ) if z > z.05 = 1.645 2.425 > 1.645; so it can be concluded that a greater proportion of Midas customers are satisfied customers in LA compared to New York. This is a hypothesis test with v = 0. Use
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(2) Is Customer Satisfaction more than 2% greater in LA? H 0 : p 1 - p 2 =.02 H A : p 1 - p 2 >.02 Select α =.05. Reject H 0 (Accept H A ) if z > z.05 = 1.645 1.679 > 1.645, so it can be concluded that customer satisfaction in LA exceeds that in New York by more than 2%. This is a hypothesis test with v ≠ 0. Use
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95% Confidence Interval for the Difference in Proportions
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Excel – Differences in Proportions =COUNTA(A2:A401) =COUNTA(B2:B201) =F1+F2 =COUNTIF(A2:A401,“YES”) =COUNTIF(B2:B201,“YES”) =F5+F6 =F5/F1 =F6/F2 =F7/F3 =(F9-F10-0)/SQRT(F11*(1-F11)*(1/F1+1/F2)) =1-NORMSDIST(E14) =(F9-F10-.02)/SQRT((F9*(1-F9)/F1)+(F10*(1-F10)/F2)) =1-NORMSDIST(E19) =(F9-F10)-NORMSINV(.975)*SQRT((F9*(1-F9)/F1)+(F10*(1-F10)/F2)) =(F9-F10)+NORMSINV(.975)*SQRT((F9*(1-F9)/F1)+(F10*(1-F10)/F2))
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Determining Sample Sizes Usual Assumptions: –Sample Sizes equal n 1 = n 2 = n –Take the “worst case scenario” for the standard deviation -- when p 1 = p 2 =.5 (unless you have reason to believe otherwise) –Use the “±” part of the confidence interval How many people need to be surveyed in each city to estimate the difference in customer satisfaction to within ± 4%?
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No Idea of the Proportions
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In a Recent Survey 90% in LA and 75% in NY Were Satisfied
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Review Hypothesis Tests for difference in proportions of the form p 1 - p 2 = 0 –By hand and by Excel Hypothesis Tests for difference in proportions of the form p 1 - p 2 = d –By hand and by Excel Confidence Intervals for difference in proportions –By hand and by Excel Estimating Sample Sizes –No idea of the proportions –Some idea of the proportions
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