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

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:

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

The test statistic is:

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

0.05 is tested at the null hypothesis:

This means that

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

This means that

Value of the test statistic:

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.

Step-by-step explanation: