Question

A poll conducted by Gallup asked American adults about vegetarianism. This problem is based on that poll. Of independent random samples of $500$men and $512$women, $20$of the men and $36$of the women said they were vegetarians. At the $5\%$significance level do the data provide sufficient level do the data provide sufficient evidence to conclude that a difference exists in the proportions of male and female vegetarians?

Part (a): Use the two-portions z-test to perform the required hypothesis test.

Part (b): Use the chi-square homogeneity test to perform the required hypothesis test.

Part (c): Compare your results in parts (a) and (b).

Part (d): Explain what principle is being illustrated.

Short answer

Part (a): We know $$\begin{gathered} H_0:p_1=p_2, \\ {H}_a:p_1\ne p_2, \\ \alpha =0.05, \\ {x}^2=-2.11 \end{gathered}$$, the critical values are $\pm 1.96$, P-value is $0.035$, where $H_0$rejects; at the $5\%$significance level, the data provide sufficient evidence to conclude that a difference exists in the proportions of male and female vegetarians.

Part (b): We know $$\begin{gathered} H_0:p_1=p_2, \\ {H}_a:p_1\ne p_2, \\ \alpha =0.05, \\ {x}^2=4.446 \end{gathered}$$, the critical values are $3.841$; $0.025<P<0.05\left(P=0.035\right)$, where $H_0$rejects; at the $5\%$significance level, the data provide sufficient evidence to conclude that a difference exists in the proportions of male and female vegetarians.

Part (c): The results of parts (a) and (b) are the same.

Part (d): The chi-square homogeneity test for comparing two population proportions and the two-tailed two proportions z-test are equivalent.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

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

Part (a) Step 2. Consider the null and alternative hypotheses.

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 $H_a:p_1\ne p_2$, 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,

Part (a) Step 3. Using the significance level.

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, $\alpha =0.05$

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

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

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

Part (b) Step 1. Consider the null and alternative hypotheses.

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,

Part (b) Step 2. Using the significance level.

Consider the above tables,

Chi-square$=4.446$,

$df=1$,

p-value$=0.035$

Using the significance level, $\alpha =0.05$

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

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

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

Part (c) Step 1. Consider parts (a) and (b).

Consider parts (a) and (b),

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

Part (d) Step 1. Consider the chi-square test.

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.