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 16 supermarkets from Region 1 had mean sales of 82.5 with a standard deviation of 6.4. A random sample of 12 supermarkets from Region 2 had a mean sales of 78.4 with a standard deviation of 6.5. Does the test marketing reveal a difference in potential mean sales per market in Region 2? Let μ1 be the mean sales per market in Region 1 and μ2 be the mean sales per market in Region 2. Use a significance level of α=0.05 for the test. Assume that the population variances are not equal and that the two populations are normally distributed.

Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.

Answer 1

the data in usual notation is for region 1

sample size n1 is 16

Solution- notation is The data in usual for begion 1 samble size n =16 Sample mean (5, ) = 82.5 Sample standard deviation (S.) = 6.4 For Region & Sample size (na) = 12 Sample mean (I) = 78.4 Samble Standard deviation (Sa)=6.5 Hypothesis Null Hypothelis, Moj H₂ = 42 Alternative Hybothelis, Hi Hi & Mg (46oo-tailed) Test statistics since the assumbtions for independent sambles 38 t-test fulfilled statistics is given by test the So, to-ta t = tn, thy-2

where san cm-1,38 + (09-1152 [ com U nitng-2 +(12-1) (6.532) sa o 16+12-2 [(16-1)16.4327 ob [61447 614.4+ 464.75] 58 41.5057 ll t = 82.5-78.4 -t 16+12-2 (41.5057) 19 4:1 t = 2.4602 Jur-505710 (41.5057/0.06asto 8833) 1.666 t= with But the critical value of t at 2=0.05 26 2.0555 is tcrict

Compawilion i so Reject Hi since tc tcritt reject Hoi and We fail to We evidence mean the sales a conclusion- do not have any that there is difference in ber market her month between two regions. е