Question

A grocery store has an average sales of $8000 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 $8300 per day. From past information, it is known that the standard deviation of the population is $1200. The p-value is

Answer 1

The p-value of the test is 0.023.

Explanation:

In this case we need to determine whether the addition of several advertising campaigns increased the sales or not.

The hypothesis can be defined as follows:

H₀: The stores average sales is $8000 per day, i.e. μ = 8000.

Hₐ: The stores average sales is more than $8000 per day, i.e. μ > 8000.

The information provided is:

`n=64\\\bar x=\$8300\\\sigma=\$1200`

As the population standard deviation is provided, we will use a z-test for single mean.

Compute the test statistic value as follows:

`z=(\bar x-\mu)/(\sigma/√(n))=(8300-8000)/(1200/√(64))=2`

The test statistic value is 2.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the two-tailed test as follows:

`p-value=P(Z>2)\\=1-P(Z<2)\\=1-0.97725\\=0.02275\\\approx 0.023`

*Use a z-table for the probability.

The p-value of the test is 0.023.