The per-store daily customer count e·the mean number of customers in a store in

Question

The per-store daily customer count e·the mean number of customers in a store in one day) for a nationwide convenience store chain that operates neary 10000 stores has been steady, at 891, for some time. To increase the customer count, the chain is considering cutting prices for coffee beverages by approximately half. To test the new initiative, the chain has reduced coffee prices in a sample of 39 stores, where customer counts have been running almost exactly at the national average of 891. After four weeks, the sample stores stabilize at a mean customer count of 932 and a standard deviation of 99. This increase seems like a substantial amount to you, but it also seems like a prety small sample. Do you think reducing coffee prices is a good strategy for increasing the mean customer count? Complete parts (a) through (c) below. a. State the null and alternative hypotheses (The population per store daly customer count mean with the reduced pnces is denoted by answer below. Choose the correct O A. Hou 891 B. How 932 H1 : 891 H 932 HH>891 >932 H1P 932 H1 : 891 C. Hou-932 b. At the 0.05 level of significance, is there evidence that reducing coffee prices is a good strategy for increasing the mean customer count? Find the test statistic and p-value for this hypothesis test. tSTAT-1 (Round to two decimal places as needed.)

Explanation / Answer

here std errror of mean =std deviation/(n)1/2 =99/(39)1/2 =15.85

therefore test statistic t =(X-mean)/std error =(932-891)/15.85=2.59

p value for right tailed test =0.0068

please revert/

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