Answer to Question #349715 in Statistics and Probability for mark jhun

Question

THE PRESIDENT OF A SERVICE UTILITY CLAIMS THAT 60 PERCENT OF HIS 500,000 CUSTOMERS ARE VERY SATISFIED WITH THE SERVICE THEY RECEIVE. TO TEST THIS CLAIM, THE LOCAL NEWSPAPER SURVEYED CUSTOMERS, USING SIMPLE RANDOM SAMPLING. AMONG THE SAMPLED CUSTOMERS, 64 PERCENT SAY THEY ARE VERY SATISFIED. BASED ON THESE FINDINGS, CAN WE REJECT THE PRESIDENT'S HYPOTHESIS THAT 64% OF THE CUSTOMERS ARE VERY SATISFIED? USE A 0.01 LEVEL OF SIGNIFICANCE.

Answer 1

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

$H_0:p=0.6$

$H_a:p\not=0.6$

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

Evidence:

Based on the information provided, the significance level is $\alpha = 0.01 ,$ and the critical value for a two-tailed test is $z_c =2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z|>2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.64-0.6}{\sqrt{\dfrac{0.6(1-0.6)}{50000}}}=57.735$$

Since it is observed that $z = 57.735>2.5758= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(Z>57.735)= 0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is different than 0.60, at the $\alpha = 0.01$ significance level.