Answer to Question #346288 in Statistics and Probability for pogi

Question

The owner of a fast-food shop claims more than 45% of the household within 5-kilometer radius of the fast-food shop have at least one member who eats at the fast-food shop at least once a week. In a sample of 250 households in the area, an investigator found that members of 180 households did so. Do these data provide sufficient evidence to support the fast-food shop owner’s claim? Use 90% level of confidence.

Answer 1

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

$H_0:p\le0.45$

$H_a:p>0.45$

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

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

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

The z-statistic is computed as follows:

$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{180/250-0.45}{\sqrt{\dfrac{0.45(1-0.45)}{250}}}=8.5812$

Since it is observed that $z = 8.5812>1.2816= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z>8.5812)=0,$ and since $p=0<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is greater than 0.45, at the $\alpha = 0.10$ significance level.