Answer to Question #264640 in Statistics and Probability for Sadie

Question

A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480 were delivered on time.

a)    Construct a 95% confidence interval for the proportion of all packages that are delivered on time by the postal service.

b)   The manager of the Postal Services claims that 98 percent of their mail are delivered on time. He wants to test the 1% significance level to determine whether the true proportion is less than 98%

                              i.           Give the null and alternative hypothesis of this test.

                            ii.           Determine the critical value(s) of this test.

                         iii.           Compute the value of the test statistic

                          iv.           State the decision rule.

                            v.           Give your decision based on the available sample evidence.

                          vi.           Hence, state your conclusion.

Answer 1

a) The sample proportion is computed as follows, based on the sample size $N = 500$

and the number of favorable cases $X = 480:$

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

The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

The corresponding confidence interval is computed as shown below:

$$CI(proportion)$$

$$=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}},\hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}})$$

$$=(0.96-1.96\sqrt{\dfrac{0.96(1-0.96)}{500}},$$

$$0.96+1.96\sqrt{\dfrac{0.96(1-0.96)}{500}})$$

$$=(0.943, 0.977)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.943 < p < 0.977,$ which indicates that we are 95% confident that the true population proportion $p$ is contained by the interval $(0.943, 0.977).$

b)

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:

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

$$=\dfrac{0.96-0.98}{\sqrt{\dfrac{0.98(1-0.98)}{500}}}=-3.1944$$

iv.

If it is observed that $z< z_c = -2.326,$ it is then concluded that the null hypothesis is rejected.

If it is observed that $p<\alpha = 0.01,$ it is then concluded that the null hypothesis is rejected.

v.Since it is observed that $z=-3.1944< -2.326=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z<-3.1944)=0.000701,$ and since $p=0.000701<0.01=\alpha,$ it is then 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.