Question

A recent survey reported that small businesses spend 24 hours a week marketing their business. A local chamber of commerce claims that small businesses in their area are not growing because these businesses are spending less than 24 hours a week on marketing. The chamber conducts a survey of 93 small businesses within their state and finds that the average amount of time spent on marketing is 23.0 hours a week. Assuming that the population standard deviation is 5.5 hours, is there sufficient evidence to support the chamber of commerce’s claim at the 0.02 level of significance?
Step 1 of 3 :
State the null and alternative hypotheses for the test. Fill in the blank below.
H0: μ=24
Ha: μ ____ 24
Step 2 of 3:
What is the test statistic?
Step 3 of 3:
Do we reject the null hypothesis? Is there sufficient or insufficient evidence?

Answer 1

Answer:

Ha: μ < 24

The test statistic is z = -1.75.

The pvalue of the test is 0.0401 > 0.02, which means that we do not reject the null hypothesis, as there is insufficient evidence.

Step-by-step explanation:

A recent survey reported that small businesses spend 24 hours a week marketing their business.

This means that the null hypothesis is:

$H_0: \mu = 24$

A local chamber of commerce claims that small businesses in their area are not growing because these businesses are spending less than 24 hours a week on marketing.

This means that the alternate hypothesis is:

$H_a: \mu < 24$

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.

24 is tested at the null hypothesis:

This means that $\mu = 24$

The chamber conducts a survey of 93 small businesses within their state and finds that the average amount of time spent on marketing is 23.0 hours a week.

This means that $n = 93, X = 23$

The population standard deviation is 5.5 hours

This means that $\sigma = 5.5$

Value of the test-statistic:

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

$z = \frac{23 - 24}{\frac{5.5}{\sqrt{93}}}$

$z = -1.75$

The test statistic is z = -1.75.

Do we reject the null hypothesis? Is there sufficient or insufficient evidence?

The pvalue of the test is the probability of finding a sample mean below 23, which is the pvalue of z = -1.75.

Looking at the z table, z = -1.75 has a pvalue of 0.0401

The pvalue of the test is 0.0401 > 0.02, which means that we do not reject the null hypothesis, as there is insufficient evidence.