Answer to Question #333448 in Statistics and Probability for Jasmine

Question

A concerned citizen group claims that 50% of the people in the city A support making beers illegal. Yoi decide to test this claim and ask a random sample of 315 out of 6000 people in the concerned city whether they support making beers illegal, 49% support this law. At 0.05 significance level, is there enough evidence to support this claim?

Answer 1

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p=0.5$

$H_1:p\not=0.5$

This corresponds to a two-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z: |z| > 1.96\}$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.49-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{315}}}\approx-0.3550$$

Since it is observed that $|z| = 0.355 \le 1.96=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p =2P(Z<-0.3550)= 0.72259,$ and since $p = 0.7226 \ge 0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is different than 0.5, at the $\alpha = 0.05$ significance level.