Question

Math533 Keller graduate university Project B

Complete the following four hypotheses, using α = 0.05 for each. a. Mean sales for all stores is at least $52,000 b. Proportion of stores with a medium market size is less than 55% c. Mean sales for promotion #3 is more than $45,000 per week d. Mean store age is greater than 8 years. 1. Using the same data set from part A, perform the hypothesis test for each speculation in order to see if there is evidence to support the manager's belief. Use the Seven Elements of a Test of Hypothesis from Section 7.1 of your text book, as well as the p-value calculation from Section 7.3, and explain your conclusion in simple terms. 2. Compute 99% confidence intervals for each of the variables described in a.-d., and interpret these intervals. 3. Write a report about the results, distilling down the results in a way that would be understandable to someone who does not know statistics. Clear explanations and interpretations are critical.

Answer 1

1.

a.

Let the population mean =$\mu$

1.Null Hypothesis(H0):

$\mu <\ $52,000$

2. Alternative Hypothesis(H1):

$\mu \geq\ $52,000$ (right tailed test)

3. Set significance level($\alpha$​​​​​​):

$\alpha =0.05$ =5%

4. Collect data:

Let the sample size, sample mean and sample standard deviation of the data (since not given in the question) be:

Sample size, n =25; Sample mean, $\bar{X}$ =52,500 and Sample std.deviation, s =4,000

5. Test statistic:

t =$(\bar{X} - \mu)/(s/\sqrt{n})$ =(52,500 - 52,000)/(4,000/$\sqrt{25}$​​​​​​) =0.625

6. Rejection region and p-value:

At 5% significance level and n-1 =24 degrees of freedom, the critical value of t for right tailed test is t_crit =1.711

So, the rejection region is as shown below:

The p-value of the test statistic of t= 0.625 with 24 degrees of freedom for one-tailed test (right tailed) is p =0.269.

7. Conclusion:

We failed to reject the null hypothesis at 0.05 significance level because the test statistic of 0.625 didnot fall in the rejection region.

Also, p-value of the test statistic is p =0.269 > 0.05 significance level. So, we failed to reject the null hypothesis.

Thus, we do not have sufficient evidence to claim that the mean sales for all stores is at least $52,000.

b.

Let the population proportion =P

1. Null Hypothesis(H0):

$P \geq 0.55$

2. Alternative Hypothesis(H1):

$P < 0.55$ (left-tailed test)

3. Set significance level:

$\alpha =0.05$

4. Collect data:

Let the Sample size, n= 36; Sample proportion, p =0.42 and so, Standard error of sample proportion, S.E(p) =$\sqrt{p(1-p)/n}$ =$\sqrt{0.42(1-0.42)/36}$ =0.0823

5. Test statistic:

Z =(p - P)/S.E = (0.42 - 0.55)/0.0823 = -1.58

6. Rejection region and p-value:

At 5% significance level, for left-tailed test, the critical value of Z is Z_crit = -1.645

So, the rejection region is as follows:

The p-value for one tailed test (left-tailed) for the test statistic of Z = -1.58 is p =0.057

7. Conclusion:

We failed to reject the null hypothesis at 0.05 significance level because the test statistic of -1.58 did not fall in the rejection region. Also, p-value of 0.057 > 0.05 significance level. So, we failed to reject the null hypothesis.

Thus, there is no sufficient evidence to claim that the proportion(P) of stores with a medium market size is less than 55%.

c. and d. are similar to a.

That is, c. and d. both are right tailed tests.