Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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

Sagot :

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:

[tex]H_0: \mu = 900[/tex]

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

[tex]H_a: \mu < 900[/tex]

The test statistic is:

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

[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] 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 [tex]\mu = 900[/tex]

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 [tex]n = 64, X= 892.50, s = 24[/tex]

Value of the test:

[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{892.50 - 900}{\frac{24}{\sqrt{64}}}[/tex]

[tex]t = -2.5[/tex]

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.

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.