Question

4. A recent study of undergraduates looked at gender differences in dieting trends. There are 181 women and 105 men who participated in the survey. The following table summarizes whether a student tried a low-fat diet or not by gender. Female Tried low-fat diet Male 35 No Yes (a) Fill in the missing cells in the table, namely, what is the number of male students who have not tried a low-fat diet and what is the number of female students who have not tried a low-fat diet? (b) Test the null hypotheses that there is no Association between gender and the likelihood of trying a low-fat diet. Give the null and alternative hypothesis, compute the test statistic and its degrees of freedom (c) Use the x table to bound your P-value. What is your conclusion at 0.057

Answer 1

$(a)\\\\ \begin{matrix} Tried~low-fat~diet & Male &Female &Totals \\ \hline Yes&35 & 8 &35+8=43 \\ No&105-35=70 &181-8=173 &70+173=243 \\ \hline Totals &105 & 181 & 43+243=286\end{matrix}\\\\ (b)~Null~hypothesis~H_0:~There~is~no~association~between~gender~and~the~likelihood~of~trying\\~a~low-fat~vs.~Alternative~hypothesis,~H_A:~H_0~is~not~true.\\\\ f_{ij}=observed~frequency~of~(i,j)th~cell;~i,j=1,2\\\\ f_{11}=35,~f_{12}=8,~f_{21}=70,~f_{22}=173.\\\\ Now,~f_{10}=43,~f_{20}=243,~f_{01}=105,~f_{02}=181,~\\N=total~frequency=286\\\\ e_{ij}=Expected~frequency~of~(i,j)th~cell=\frac{f_{i0}f_{0j}}{N};~i,j=1,2\\\\ then~e_{11}=(43*105)/286=15.7867,~e_{12}=(43*181)/286=27.2133,\\\\ e_{21}=(243*105)/286=89.2133,~e_{22}=(243*181)/286=153.7867.\\\\ Value~of~\chi^2=\sum_{i=1}^2\sum_{j=1}^2\frac{(f_{ij}-e_{ij})^2}{e_{ij}} =43.4870\\\\ Degrees~of~freedom=(2-1)(2-1)=1\\\\$

$(c)~P-value=P(\chi^2>43.4870|\chi^2\sim \chi^2_1)<0.05~so~we~reject~H_0~at~5\%~level~of~significance~hence\\ there~is~a~significance~association~between~gender~and~the~likelihood~of~trying~a~low-fat.$