Question

The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a diference in the proportion of males and temales who use the local routes. She takes a sample of a day's riders and finds the tolowing Female 36 25 Male Loca 35 Non-Loca She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

Answer 1

$Observed~frequencies: \\\begin{matrix} & Male &Female&Totals \\ Local &35=f_{11} & 36=f_{12}&71=f_{10}\\ Non-local&33=f_{21} & 25=f_{22}&58=f_{20}\\ Totals&68=f_{01}&61=f_{02}&129=N \end{matrix}\\ where,~f_{ij}=observed~frequency~of~(i,j)^{th}~cell\\ f_{i0}=f_{i1}+f_{i2},~i=1,2,~f_{0j}=f_{1j}+f_{2j},~j=1,2\\ N=f_{11}+f_{12}+f_{21}+f_{22}\\ Expected~frequencies:~e_{ij}=\frac{f_{i0}f_{0j}}{N},~i.j=1,2\\ \begin{matrix} & Male &Female\\ Local &(71*68)/129=37.4264= e_{11} & (71*61)/129=33.5736=e_{12}\\ Non-local&(58*68)/129= 30.5736=e_{21} & (58*61)/129=27.4264=e_{22}\\ \end{matrix}\\ \chi^2~test~statistic=\sum_{i=1}^2\sum_{j=1}^2\frac{(f_{ij}-e_{ij})^2}{e_{ij}}\\ Value~of~\chi^2~test~statistic=(35-37.4264)^2/37.4264+(36-33.5736)^2/33.5736+(33-30.5736)^2/30.5736+(25-27.4264)^2/27.4264=0.7399\\$

$p-value=P(\chi^2>0.7399|\chi^2\sim \chi^2_1)=0.3897>0.05~so~we~fail~to~reject~null~hypothesis\\ at~5\%~level~of~significance~and~conclude~that~there~is~insignificant~difference~in~the\\ proportion~of~males~and~females~who~use~the~local~routes.$