Question

A company hopes to improve customer satisfaction, setting a goal of less than 5% negative comments. A random survey of 850 customers found only 34 with complaints. Does this provide evidence at the 10% significance level that the company has reached its goal of decreasing the percentage of complaints

Answer 1

Answer:

The p-value of the test is 0.0901 < 0.1, which means that this provides evidence at the 10% significance level that the company has reached its goal of decreasing the percentage of complaints.

Step-by-step explanation:

A company hopes to improve customer satisfaction, setting a goal of less than 5% negative comments.

At the null hypothesis, we test if the proportion of negative comments is of at least 5%, that is:

$H_0: p \geq 0.05$

At the alternative hypothesis, we test if this proportion is less than 0.05, that is:

$H_1: p < 0.05$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.05 is tested at the null hypothesis:

This means that $\mu = 0.05, \sigma = \sqrt{0.05*0.95}$

A random survey of 850 customers found only 34 with complaints.

This means that $n = 850, X = \frac{34}{850} = 0.04$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.04 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{850}}}$

$z = -1.34$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.04, which is the p-value of z = -1.34.

Looking at the z-table, z = -1.34 has a p-value of 0.0901.

The p-value of the test is 0.0901 < 0.1, which means that this provides evidence at the 10% significance level that the company has reached its goal of decreasing the percentage of complaints.