Question

Consider a company that stocks it's CD players in stores. It wants to determine if the location on where their product is placed in the store matters for sales generated. They contract with the retailer to place their product in the “front of the store” or in the “aisle of the store.” The collected data on the sales of their products separated by the location in the store. Test whether there is a significant difference in the mean sales across the two different locations in the store.

Using software:

a. Generate summary statistics (central tendency and variability measures) for the two samples and briefly summarize what they say.

b. Construct a 95% confidence interval of the difference between the two population mean delivery times and interpret it.

c. Finally, are there any other factors, besides the way the store is classified, that could possibly influence delivery times (identify at least 2)? Include a brief explanation of each.

Front of the store		Aisle of the store
224		192
189		236
248		164
285		154
273		189
190		220
243		261
215		186
280		219
317		202

Answer 1

In this case study, we want to determine sales of the CD players is the same or different by location

in the “front of the store” or in the “aisle of the store.”

a) Summary Statistics:

For variable Front_of_the_store

Min. 1st Qu. Median Mean 3rd Qu. Max.
189.0 217.2 245.5 246.4 278.2 317.0

The mean and Median value is nearly equal means variable would be symmetric.

Standard deviation = 42.54201

Min. 1st Qu. Median Mean 3rd Qu. Max.
154.0 186.8 197.0 202.3 219.8 261.0

The mean and Median value is not equal means variable could not symmetric.

Standard deviation =  32.52708

The mean of first sample is higher than second means sales of CD player is more in front of store.

b) 95 % Confidence interval for mean difference is,

Lower Limit = M₁ - M₂ -(t_CL)()
Upper Limit = M₁ - M₂ +(t_CL)()

where M1 and M2 are the means of the above two samples and S is pooled variance and t test statistics.

By using R we can compute easily this value,

95 percent confidence interval:

( 8.345485, 79.854515)

c) Test Hypothesis:

H0: The sales of CD player is the same in the “front of the store” or in the “aisle of the store.”

against,

H1: The sales of CD player is not the same in the “front of the store” or in the “aisle of the store.”

Test Statistic:

We can use t-test here,

t = 2.6041, df = 16.843, p-value = 0.01862

alternative hypothesis: true difference in means is not equal to 0

Decision Rule:

If the p-value is larger than the level of significance then accept the null hypothesis.

In above example, the p-value is less than 0.05 level of significance so we reject null hypothesis here

so the sales of CD player is not the same in the “front of the store” or in the “aisle of the store.”

The other factor may distance because customers are attracted to store nearly but if it outside the no one see to it.

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