Answer to Question #336023 in Statistics and Probability for Joseph

Question

a school administrator claims that less than 50% of the students of the school are dissatisfied by the community cafeteria service. Test this claim by using sample data obtained from a survey of 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:

$H_0:p\ge0.50$

$H_a:p<0.50$

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

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.54-0.50}{\sqrt{\dfrac{0.50(1-0.50)}{500}}}\approx1.7889$$

The p-value is $p =P(Z<1.7889)= 0.963185$

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 right-tailed test is $R = \{z: z<-1.6449\}.$

Since it is observed that $z = 1.7889 \ge-1.6449= z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p =0.963185,$ and since $p = 0.963185 \ge 0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is less than 0.50, at the $\alpha = 0.05$ significance level.