Question

The marketing manager of a firm that produces laundry products decides to test market a new laundry product in each of the firm's two sales regions. He wants to determine whether there will be a difference in mean sales per market per month between the two regions. A random sample of 12 supermarkets from Region 1 had mean sales of 84.1 with a standard deviation of 7.6. A random sample of 16 supermarkets from Region 2 had a mean sales of 76.3 with a standard deviation of 6.9 . Does the test marketing reveal a difference in potential mean sales per market in Region 2? Let LaTeX: \muμ1 be the mean sales per market in Region 1 and LaTeX: \muμ2 be the mean sales per market in Region 2. Use a significance level of 0.05 for the test. Assume that the population variances are not equal and that the two populations are normally distributed.

Solution:

Step 1. State the null and alternative hypotheses for the test. Answer: ____________________

Step 2. Compute the value of the t- test statistic. Round your answer to three decimal places. Answer: ____________________

Step 3. Determine the decision rule for rejecting the null hypothesis H0 . Round your answer to three decimal places. Answer: ____________________

Step 4. State the test's conclusion. A) Reject Null Hypothesis B) Fail to Reject Null Hypothesis

Answer 1

T-test for two Means – Unknown Population Standard Deviations - Unequal Variance

The following information about the samples has been provided: a. Sample Means : Xˉ1​=84.1 and Xˉ2​=76.3 b. Sample Standard deviation: s1=7.6 and s2=6.9 c. Sample size: n1=12 and n2=16 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ1 =μ2 Ha: μ1 ≠μ2 This corresponds to a Two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. The degrees of freedom Assuming that the population variances are unequal, the degrees of freedom are given by (2)Test Statistics The t-statistic is computed as follows: (3a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is 22.4989. Therefore the critical value for this Two-tailed test is tc​=2.0739. This can be found by either using excel or the t distribution table. (3b) Rejection Region The rejection region for this Two-tailed test is |t|>2.0739 i.e. t>2.0739 or t<-2.0739 (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.0106 The Decision about the null hypothesis (a) Using traditional method Since it is observed that |t|=2.7948 > tc​=2.0739, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0106, and since p=0.0106≤0.05, it is concluded that the null hypothesis is rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ1​ is different than μ2, at the 0.05 significance level.

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