Question

Sure Value Convenience Store Case Study:

You work in the corporate office for a nationwide convenient store franchise that operates nearly 10,000 stores. The per-store daily customer count (i.e., the mean number of customers in a store in one day) has been steady, at 900, for some time. To increase the customer count, the chain is considering cutting prices for coffee beverages. The question to be determined is how much to cut prices to increase the daily customer count without reducing the gross margin on coffee sales too much. You decide to carry out an experiment in a sample of 24 stores where customer counts have been running almost exactly at the national average of 900. In 6 of the stores, the price of small coffee will now be $0.59, in 6 stores the price of a small coffee will now be $0.69, in 6 stores, the price of a small coffee will now be $0.79, and in 6 stores, the price of a small coffee will now be $0.89. After four weeks of selling the coffee at the new price, the daily customer counts in the stores were recorded and stored as follows...

$0.59 Sales	$0.69 Sales	$0.79 Sales	$0.89 Sales
964	953	942	920
972	950	937	918
962	959	945	925
968	955	948	919
975	960	945	915
960	954	941	906

1) Analyze the data and determine whether there is evidence of a difference in the daily customer count, based on the price of a small coffee.

2) If appropriate, determine which mean prices differ in daily customer counts.

3) What price do you recommend for a small coffee?

Response should be computer-generated, such as in Word or Excel, not a photo of handwritten work. Preferably excel if at all possible. It should follow a basic memo format with introduction, analysis, and conclusions.

Answer 1

1.

Null hypothesis, H₀: There is no difference in the average daily customer count, based on the price of a small coffee.

vs.

Alternative hypothesis, H_a: H₀ is not true.

$0.59 Sales	$0.69 Sales	$0.79 Sales	$0.89 Sales		Anova: Single Factor
964	953	942	920
972	950	937	918		SUMMARY
962	959	945	925		Groups	Count	Sum	Average	Variance
968	955	948	919		$0.59 Sales	6	5801	966.8333	34.56667
975	960	945	915		$0.69 Sales	6	5731	955.1667	14.16667
960	954	941	906		$0.79 Sales	6	5658	943	14.8
					$0.89 Sales	6	5503	917.1667	40.56667
					ANOVA
					Source of Variation	SS	df	MS	F	P-value	F crit
					Between Groups	8145.458	3	2715.153	104.3286	2.2E-12	3.098391
					Within Groups	520.5	20	26.025
					Total	8665.958	23

Since p-value of F test=2.2x10^-12<0.05 so there is sufficient evidence to conclude that there is a significant difference in the daily customer count, based on the price of a small coffee.

2. Since null hypothesis is rejected so we use Fisher's LSD method.

A=mean counts of Group named $0.59 Sales=966.8333

B=mean counts of Group named $0.69 Sales=955.1667

C=mean counts of Group named $0.79 Sales=943

D=mean counts of Group named $0.89 Sales=917.1667

Hence all mean prices are differ from each other in daily customer counts.

3. We recommend $0.59 for a small coffee since under this price average customer counts are highest.