91Ó°ÊÓ

Consider the given question,

The given independent random samples of $500$ men and$512$women.

The test hypothesis are given below,

Null hypothesis is $H_0:p_1=p_2$, that is the data do not provide sufficient evidence to conclude that a difference exists in the proportions of male and female vegetarians.

Alternative hypothesis is $H_a:p_1≠p_2$, that is the data provide sufficient evidence to conclude that a difference exists in the proportions of male and female vegetarians.

On finding the value of test statistics and P-value,

Difference$=p\left(1\right)-p\left(2\right)$

Estimate for difference is $-0.0303125$

$95\%$CI for difference is $\left(-0.0583388,-0.00228617\right)$.

Test for difference, $x^2=-2.11$

Fisher's exact test,$p-value=0.035$

Using the significance level, $\mathrm{Î\pm }=0.05$

Here, P-value is lesser than the level of significance. That is $P-value\left(=0.035\right)<\mathrm{Î\pm }\left(=0.05\right)$.

Therefore, by the rejection rule, it can be concluded that there is evidence to reject the null hypothesis $\left(H_0\right)$at $\mathrm{Î\pm }=0.05$.

Thus, the data provide sufficient evidence to conclude that a difference exists in the proportions of male and female vegetarians.

The test hypothesis are given below,

Null hypothesis:$H_0$, denotes the distribution of vegetarian preference is same for men and women.

Alternative hypothesis: $H_1$, denotes the distribution of vegetarian preference is same for men and women.

On finding the value of test statistics and P-value,

Consider the above tables,

Chi-square$=4.446$,

$df=1$,

p-value$=0.035$

Using the significance level, $\mathrm{Î\pm }=0.05$

Here, P-value is lesser than the level of significance. That is $P-value\left(=0.035\right)<\mathrm{Î\pm }\left(=0.05\right)$.

Therefore, by the rejection rule, it can be concluded that there is evidence to reject the null hypothesis $\left(H_0\right)$at $\mathrm{Î\pm }=0.05$.

Thus, the distribution of vegetarian preference is same for men and women.

Consider parts (a) and (b),

It is clear that the conclusion and P-value is exactly same for both the methods.

The principle is being illustrated in the situation is that the chi-square test for homogeneity is same as the two proportions z-test for two tailed.

As there are only two populations are considered.