Question

2) A politician claims that she will receive 60% of the votes in the upcoming election. Of a random sample of 200 voters, there were 100 who will surely vote for her. Test the politician’s assertion at the 0.05 level of significance.

Answer 1

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

$H_0:p=0.60$

$H_a:p\not=0.60$

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:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{100/200-0.6}{\sqrt{\dfrac{0.6(1-0.6)}{200}}}=-2.8284$$

Since it is observed that $|z|=2.8284 \ge1.96= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(Z<-2.8284)= 0.004678,$ and since $p=0.004678<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is different than 0.60, at the $\alpha = 0.05$ significance level.