Question #276611

(7 points) There is a popular story (among data miners) that there is a correlation between men buying diapers and buying beer while shopping. A student tests this theory by surveying 131 male shoppers as they left a grocery store. The results are summarized in the contingency table. Test for a dependent relationship between buying beer and buying diapers. Conduct this test at the 0.10 significance level.

Bought Diapers Did Not Buy Diapers Totals

Beer 7 46 53

No Beer 11 67 78

Totals 18 113 131

(a) Find the expected frequencies.

Bought Diapers Did Not Buy Diapers Beer

No Beer

(b) Find the test statistic.

(c) Find the critical value.


1
Expert's answer
2021-12-07T13:11:59-0500

 contingency table

diapersnot diapersbeer746not beer1167\def\arraystretch{1.5} \begin{array}{c:c:c} & diapers & not\space diapers\\ \hline beer & 7 & 46 \\ \hdashline not \space beer & 11 & 67 \end{array}

a) Find the expected frequencies.

diapersnot diapersbeer(7+11)(7+46)131=7.282(46+7)(46+67)131=45.718not beer(11+7)(11+67)131=10.718(67+11)(67+46)131=67.282\def\arraystretch{1.5} \begin{array}{c:c:c} & diapers & not\space diapers\\ \hline beer & \frac{(7+11)\cdot (7+46)}{131}=7.282 & \frac{(46+7)\cdot(46+67)}{131} =45.718\\ \hdashline not \space beer & \frac{(11+7)\cdot (11+67)}{131}=10.718 & \frac{(67+11)\cdot (67+46)}{131}=67.282 \end{array}


(b) Find the test statistic

χ2=(77.282)27.282+(4645.718)245.718+(1110.718)210.718+(6767.282)267.282=\chi^{2*}=\frac{(7-7.282)^2}{7.282}+\frac{(46-45.718)^2}{45.718}+\frac{(11-10.718)^2}{10.718}+\frac{(67-67.282)^2}{67.282}=0.021

(c) Find the critical value

Degrees for freedom DF=(21)(21)=1(2-1)\cdot(2-1)=1

Critical value

χkr2=qchisq(10.1,1)=2.706\chi^2_{kr}=qchisq(1-0.1,1)=2.706

χ2<χkr2\chi^{2*}<\chi^2_{kr} then H0 hypothesis of independence beer and diapers purchases not rejected on level of confidence 1-0,1=0.9


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