Question

A bank manager estimates that 75% of the customers wait less than ten minutes in the

queue at the bank. To look into this estimation, a customer group conducted a survey

with a sample of 90 customers. It was found that 15% of the customers waited at least

ten minutes in the queue. At the level of significance = 0.01, is there enough

evidence to conclude that more than 75% of the customers wait less than ten minutes

in the queue at the bank?​

Answer 1

Testing the hypothesis, it is found that since the p-value of the test is of 0.0143 > 0.01, there is not enough evidence to conclude that more than 75% of the customers wait less than ten minutes  in the queue at the bank.

At the null hypothesis, it is tested if at most 75% of the customers wait less than ten minutes  in the queue at the bank, that is:

$H_0: p \leq 0.75$

At the alternative hypothesis, it is tested if the proportion is of more than 75%, that is:

$H_1: p > 0.75$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.
• p is the proportion tested at the null hypothesis.
• n is the sample size.

For this problem, the parameters are: $p = 0.75, n = 90, \overline{p} = 1 - 0.15 = 0.85$

Hence, the value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.85 - 0.75}{\sqrt{\frac{0.75(0.25)}{90}}}$

$z = 2.19$

The p-value of the test is the probability of finding a sample proportion above 0.85, which is 1 subtracted by the p-value of z = 2.19.

Looking at the z-table, z = 2.19 has a p-value of 0.9857.

1 - 0.9857 = 0.0143.

Since the p-value of the test is of 0.0143 > 0.01, there is not enough evidence to conclude that more than 75% of the customers wait less than ten minutes  in the queue at the bank.

A similar problem is given at https://brainly.com/question/25600813