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:
<em>H₀</em>: The stores average sales is $8000 per day, i.e. <em>μ</em> = 8000.
<em>Hₐ</em>: The stores average sales is more than $8000 per day, i.e. <em>μ</em> > 8000.
The information provided is:
![n=64\\\bar x=\$8300\\\sigma=\$1200](https://tex.z-dn.net/?f=n%3D64%5C%5C%5Cbar%20x%3D%5C%248300%5C%5C%5Csigma%3D%5C%241200)
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](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Cbar%20x-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B8300-8000%7D%7B1200%2F%5Csqrt%7B64%7D%7D%3D2)
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](https://tex.z-dn.net/?f=p-value%3DP%28Z%3E2%29%5C%5C%3D1-P%28Z%3C2%29%5C%5C%3D1-0.97725%5C%5C%3D0.02275%5C%5C%5Capprox%200.023)
*Use a z-table for the probability.
The p-value of the test is 0.023.