91影视

Need to find whether there is sufficient evidence to reject the distributor's claim.

Given counts:

The expected counts are the product of the column and row total, divided by the sample size of n=107 :

The value of the test statistic is thus:

${蠂}^2=\frac{(56-59.4815{)}^2}{59.4815}+鈥�+\frac{(16-24.3399{)}^2}{24.3399}=13.305$

$0.025<P<0.05$

$P<0.05鈬�\text{Reject}{H}_0$

There is sufficient evidence to support the claim of an association.

Need to find Which cell in the table contributes most to the relationship in part (a)

The expected counts are the product of the column and row total, divided by the sample size of n=459

The value of the test statistic is thus:

$$\begin{gathered} {蠂}^2=\frac{(56-59.4815{)}^2}{59.4815}+鈥�+\frac{(16-24.3399{)}^2}{24.3399} \\ =0.204+0.186+0.044+0.092+1.177+0.700+0.641+1.693+2.704+1.865+1.141+2.858 \\ =13.305 \end{gathered}$$

The highest contribution is$2.858$which is of the "Heavy"-"University" group.