# Soln to Cpt 11 problems - Solutions to Chapter 11 Problems...

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Solutions to Chapter 11 Problems11.2(a)Observed FreqExpected FreqObserved FreqExpected FreqTotal Obs, Row 12025302550chi-sq contrib= 1.00chi-sq contrib= 1.00Observed FreqExpected FreqObserved FreqExpected FreqTotal Obs, Row 23025202550chi-sq contrib= 1.00chi-sq contrib= 1.00Total Obs, Col 1Total Obs, Col 2GRAND TOTAL5050100(b)Decision rule: If 2χ> 3.841, reject H0.Test statistic: χ2= ∑(f0fe)2fe= 1.00 + 1.00 + 1.00 + 1.00 = 4.00Decision: Since χ2calc= 4.00 is above the critical value of 3.841, reject H0.11.8(a)H0: p1= p2H1: p1p2Decision rule: df= 1. If χ2> 3.841, reject H0.Test statistic: χ2= 33.3333Decision: Since χ2calc= 33.3333 is above the upper critical bound of 3.841, reject H0. There is enough evidence to conclude that there is a significant difference between the proportion of males and females who place more importance on brand names today than a few years ago.(b)p value is virtually 0. The probability of obtaining a test statistic as large as 33.3333 or larger when the null hypothesis is true is virtually 0.11.12(a)PHStat output:Chi-Square TestObserved FrequenciesDrug RegimenResultABCDTotalSevere7147857548203073Not Severe963095439042955737772Total103441032897961037740845Expected FrequenciesDrug RegimenResultABCDTotalSevere778.2375777.0338737.0084780.72033073Not Severe9565.7629550.9669058.9929596.2837772Total103441032897961037740845DataLevel of Significance0.05Number of Rows2Number of Columns4Degrees of Freedom3ResultsCritical Value7.81472Chi-Square Test Statistic8.38264
p-Value0.03873Reject the null hypothesis0:H
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