Answer to Question #346593 in Statistics and Probability for John Lloyd

Question

A politician claims that he will get at least 70% of the votes. Out of 300 randomly sampled registered voters, 200 said they will vote for the said politician. Test the claim using 0.10 level of significance.

Answer 1

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

$H_0:p\ge0.7$

$H_a:p<0.7$

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

Evidence:

Based on the information provided, the significance level is $\alpha = 0.10 ,$ and the critical value for a left-tailed test is $z_c = -1.2816.$

The rejection region for this left-tailed test is $R = \{z: z < -1.2816\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{200/300-0.7}{\sqrt{\dfrac{0.7(1-0.7)}{300}}}=-1.26$$

Since it is observed that $z = -1.26>-1.2816= z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(Z<-1.26)= 0.103835,$ and since $p=0.103835>0.10=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is less than 0.7, at the $\alpha = 0.10$ significance level.