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.(Provide mathematical steps and explanations in detail to receive full credit.) The researcher would like to test the following hypothesis ????: ?? ? 8,000 ????: ?? > 8,000 (a) Calculate the value of the test statistic. Let ?=0.05 (b) What is the conclusion based on the critical value approach? Let ?=0.05 (c) What is the conclusion based on the P-VALUE value approach? Let ?=0.05

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:  [tex]n=64\\\bar x=\$8300\\\sigma=\$1200[/tex]As the population standard deviation is provided, we will use a z-test for single mean. Compute the test statistic value as follows:  [tex]z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{8300-8000}{1200/\sqrt{64}}=2[/tex]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:  [tex]p-value=P(Z>2)\\=1-P(Z<2)\\=1-0.97725\\=0.02275\\\approx 0.023[/tex]*Use a z-table for the probability. The p-value of the test is 0.023.