Question

The manager of the Postal Services claims that 98 percent of their mail are delivered on time. He wants to test at the 1% significance level to determine whether the true proportion is less than 98 percent. i. Give the null and alternative hypothesis of this test. [2] ii. Determine the critical value(s) of this test. [2] iii. Compute the value of the test statistic. [2] iv. State the decision rule. [1] v. Give your decision based on the available sample evidence. [1] vi. Hence, state your conclusion

Answer 1

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

$H_0:p\geq0.98$

$H_1:p<0.98$

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

ii. Based on the information provided, the significance level is $\alpha = 0.01 ,$ and the critical value for a left-tailed test is $z_c = -2.3263.$

The rejection region for this left-tailed test is $R = \{z: z < -2.3263\}.$

iii. The z-statistic is computed as follows:

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

iv. If it is observed that $z < z_c = -2.3263,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z<z),$ and if $p<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

v. Let $N=500, X=480.$ Then

$$\hat{p}=\dfrac{480}{500}=0.96$$

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.96-0.98}{\sqrt{\dfrac{0.98(1-0.98)}{500}}}$$ $$=-3.1944$$

Since it is observed that $z =-3.1944 < z_c = -2.3263,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z<-3.1944)=0.0007,$ and if $p=0.0007<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

vi. Therefore, there is enough evidence to claim that the population proportion $p$ is less than 0.98, at the $\alpha = 0.01$ significance level.