Answer to Question #259936 in Statistics and Probability for Jenelle

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.