State the hypothesis that you are checking. Summarize your results.

Question

8.      The CEO of a major chain of stores claims that a particular store’s average daily sales is significantly lower than the average daily sales of all the branch stores. The average daily sales at all the branch stores is $10,000. The average of 30 days of sales for the store in question is $9750 with a standard deviation of $500. Test the CEO’s claim at the 0.05 alpha level. Copy and Paste information from the ESP4 sheet, including the p-value. State the hypothesis that you are checking. Summarize your results.

Answer 1

Fun question!

Let's start with a statement of our null and alternative hypotheses:

Null hypothesis:

H_0: the average daily sales of a particular store is equal to the average daily sales of all branch stores.

Alternative hypothesis:

H_a: the average daily sales of a particular store is less than the average daily sales of all branch stores.

Next, grab the information provided to us in the question: (define what we know)

Average daily sales at all branch locations (population mean) mu = 10,000

Average daily sales at particular location (sample mean) x-bar = 9,750

Standard deviation of average daily sales at particular location s = 500

Number of Days that we collect data (observations) n = 30

Alpha = 0.05

Next, we determine the proper test:

Because we are not given the population variance, we will use the Student's t distribution to calculate a test statistic. The formula for our test statistic will be:

(test statistic) = (sample_mean - population_mean) / (sample_standard_deviation / sqrt(n))

Substituting our numbers into the equation:

-2.739 = (9,750 - 10,000) / (500 / sqrt(30))

Next, find the Student's t distribution critical values table, calculate the degrees of freedom (df = n-1) (df=29). Use the far-left column to find df = 29. Next, find the column for a single-tailed test that corresponds to an alpha of 0.05. Notice the value there is -1.699 - this is the critical value for t. We compare the test statistic to the critical value, finding -2.739 is more extreme than -1.699. Therefore, we reject the null hypothesis and determine that the particular store has lower average daily sales than the population.

To find the p value, plug our test statistic into the technology of your choice, with a degrees of freedom = 29, to find a p value of .005214.

Here's an article on calculating p value for a Student's t test statistic:

https://www.statology.org/p-value-from-t-score-excel/

Hope this helps!