Problem Statement
Is reducing coffee prices good strategy for increasing the mean customer count in the coffee restaurants?
Method
A sample if 34 stores is selected where customers counts have been running with small and big stores. The calculated descriptive statistics are a shown in the table below:
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|
Descriptive Statistics |
|
|
Mean |
925.79 |
|
Standard Error |
12.21 |
|
Median |
908.50 |
|
Mode |
867.00 |
|
Standard Deviation |
71.21 |
|
Sample Variance |
5070.65 |
|
Kurtosis |
0.31 |
|
Skewness |
0.37 |
|
Range |
315.00 |
|
Minimum |
787.00 |
|
Maximum |
1102.00 |
|
Sum |
31477.00 |
|
Count |
34 |
The value of the mean=925.76, the median =908.50, the maximum number of stores is 1102 while the minimum is 787 stores. The standard deviation is 71.21 with a sample variance of 5070.65.
The scatterplot to show the plot of the number of customers is as shown below:
The plot shows that the number of customers that visited the 34 stores is between 750 and 1052 customers. Further, to demonstrate the distribution of the average number of customers, a histogram is shown below:
Assumptions
The analysis assumed that the stores under the study is in Texas and New Mexico. In addition, it’s assumed that the cut prices of the stores will lead to an increase in the number of customers.
Finally, its assumed that the main reason behind making the decisions increasing the average number of customers.
Analysis
The hypothesis tested in the analysis is that; H 0 : μ ≤ 900
H 1 : μ > 900
The type of test conducted to test the hypothesis is a one sample t-test at a significance level of 0.05.
|
t-Test: One-Sample |
|
|
Number of Customers |
|
|
Mean |
925.7941176 |
|
Variance |
5070.653298 |
|
Observations |
34 |
|
Hypothesized Mean |
900 |
|
df |
33 |
|
t Stat |
2.112166627 |
|
P(T<=t) one-tail |
0.021162767 |
|
t Critical one-tail |
1.692360309 |
|
P(T<=t) two-tail |
0.042325535 |
|
t Critical two-tail |
2.034515297 |
The t-test output above shows that the average number of customers obtained is 926. Comparing the hypothesized value of 900 with the average number of customers, the t-statistics is 2.1122 with a p-value of p=0.0212 for a one tailed hypothesis. Since the p-value is less than the significance value, the null hypothesis is rejected that theirs is no difference between the means and conclude that a significant difference does exist. In other words, we can conclude that the change in the price is effective as it will lead to an increase in the number of customers in the stores.
Recommendation
Based on the results from the hypothesis tested, it can be recommended to the management of the stores that reducing the prices of the stores is effective in achieving the average of more than 900 customers. They should thus implement the decrease of prices as earlier proposed.
References
Badenes-Ribera, L., Frías-Navarro, D., Monterde-i-Bort, H., & Pascual-Soler, M. (2015). Interpretation of the p value: A national survey study in academic psychologists from Spain. Psicothema , 27 (3), 290-295.
Croft, D. P., Madden, J. R., Franks, D. W., & James, R. (2011). Hypothesis testing in animal social networks. Trends in ecology & evolution , 26 (10), 502-507.
Ross, A., & Willson, V. L. (2017). One-sample T-test. In Basic and Advanced Statistical Tests (pp. 9-12). Brill Sense.
