Question

Based on past experience, the daily revenue, X, for a coffee shop in a city is a normally distributed random variable with a mean daily revenue of $900. The first co-owner of the shop feels that the shop is known throughout the city and as such they could reduce the amount of money spent on advertising without causing a decrease in the population mean daily revenue. The second co-owner is concerned that a reduction in advertising would in fact result in a decrease in the population mean daily revenue. To settle the issue, the first co-owner decides to decrease the advertising dollars for a period of time, during which she will select 64 days at random and compute the sample mean daily revenue. She will then perform a hypothesis test to determine if the sample mean daily revenue provides significant evidence to conclude that the population mean daily revenue is less than $900 when the advertising dollars are reduced. The first co-owner observes a sample mean daily revenue of $892.50 with a sample standard deviation of $24 when the advertising dollars have been reduced. What is the p-value for her test

Answer 1

Answer:

The p-value for her test is 0.0075.

Step-by-step explanation:

She will then perform a hypothesis test to determine if the sample mean daily revenue provides significant evidence to conclude that the population mean daily revenue is less than $900 when the advertising dollars are reduced.

At the null hypothesis, we test that the mean is less than 900, that is:

$H_0: \mu = 900$

At the alternate hypothesis, we test that the mean is less than 900, that is:

$H_a: \mu < 900$

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

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

900 is tested at the null hypothesis:

This means that $\mu = 900$

Sample of 64 days. The first co-owner observes a sample mean daily revenue of $892.50 with a sample standard deviation of $24 when the advertising dollars have been reduced.

This means that $n = 64, X= 892.50, s = 24$

Value of the test:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{892.50 - 900}{\frac{24}{\sqrt{64}}}$

$t = -2.5$

What is the p-value for her test?

The pvalue of the test is the probability of a sample mean below 892.5, which is a left-tailed test for t = -2.5 with 64 - 1 = 63 degrees of freedom.

Using a calculator, the p-value for her test is 0.0075.