Exhibit 9-9
The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8,300 per day. From past information, it is known that the standard deviation of the population is $1,200.
-Refer to Exhibit 9-9. The p-value is
A)2.00
B)0.9772
C)0.0228
D)0.5475

Question

Quizplus · Accepted Answer

To determine the p-value, we need to conduct a one-sample t-test with the null hypothesis that the population mean is equal to $8,000 and the alternative hypothesis that the population mean is greater than $8,000 (assuming that the advertising campaigns have increased sales).

The test statistic can be calculated as:

t = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
t = (8,300 - 8,000) / (1,200 / sqrt(64))
t = 2.67

Using a t-distribution table with degrees of freedom = 63 (sample size minus one), we find that the p-value associated with a t-value of 2.67 and a one-tailed test is 0.0228.

Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the advertising campaigns have increased sales.

Exhibit 9-9 the Sales of a Grocery Store Had an Average of Average