Question

Directions: Use the traditional method and p-value method in solving the problem on population proportion. Problem: A school administrator claims that less than 50% of the students of the school are dissatisfied by the community cafeteria service. Test 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 = 0.05 

Check the assumptions. Is the sample size large enough for the Central Limit Theorem (CLT) to apply? With n= ________, the Central Limit Theorem applies. 

Answer 1

When the sample size is large the sample proportion is normally distributed.

$$np=500(0.5)=250>30$$

$$nq=500(0.5)=250>30$$

With $n=500,$ the Central Limit Theorem applies.

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.

Evidence:

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:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.54-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{500}}}=3.5777$$

Since it is observed that $z =3.5777 \ge-1.6449= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z<3.5777)= 0.999827,$ and since $p=0.999827>0.05=\alpha,$ 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 0.50, at the $\alpha = 0.05$ significance level.