Answer to Question #204928 in Statistics and Probability for Cashper

Question

a school administrator claims that less than 50% of the students of the school are dissatisfied by community cafeteria service. test this claim by using sample data obtained from a survey o 500 students of the school where 54% indicated their dissatisfaction of the community cafeteria service. use a =0.05

Answer 1

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

$H_0: p\geq0.5$

$H_1:p<0.5$

This corresponds to a left-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 left-tailed test is $z_c=-1.6449.$

The rejection region for this left-tailed test is $R=\{z: z<-1.6449\}$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.54-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{500}}}\approx1.7889$$

Since it is observed that $z=1.7889>-1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

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

Using the P-value approach: The p-value is $p=P(z<1.7889)=0.9632,$ and since $p=0.9632>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

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