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

Answer:

The p-value of the test is 0.023.

Step-by-step 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:

 $z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{8300-8000}{1200/\sqrt{64}}=2$

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.