Question

A grocery store wants to know which system is better for decreasing the amount of time that their customers wait in checkout lines. Is there a significant difference in the average time customers wait in line using the cashier operated checkout and the self-checkout? The data was collected by recording the amount of time it takes to check out 20 customers that enter the cashier checkout line and 20 customers that enter the self-checkout line. The number of minutes from when the customer enters the checkout line completes the check out process and exits the line.

Cashier Electronic
4.21 9.66
5.55 5.9
3.02 8.02
5.13 5.79
4.77 8.73
2.34 3.82
3.54 8.01
3.2 8.35
4.5 10.49
6.1 6.68
0.38 5.64
5.12 4.08
4.26    6.71
6.19 9.91
3.79 5.47
4.25 4.06
6.32 8.65
4.45 3.5
4.26 5.58
3.46 7.41

1. Report how you analyzed the data using a two-sample t-test analysis technique

2. Include the output and any calculations of the analysis you performed on Excel (screenshot or picture).

3.. Discuss the results of your data analysis.

4. Discuss the limitation(s) of your data analysis.

5. Recommend a course of action based on your results.

Answer 1

SOLUTION 1: LET US ASSUME THAT THE TWO POPULATION HAVE EQUAL VARIANCES AND DATA ARE NORMALLY DISTRIBUTED.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ =μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=38. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc​=2.024, α=0.05 and df=38.

The rejection region for this two-tailed test is R={t:∣t∣>2.024}.

SOLUTION 3: Decision about the null hypothesis

Since it is observed that ∣t∣=4.533>tc​=2.024, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0001, and since p=0.0001<0.05, it is concluded that the null hypothesis is rejected.

Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is different than μ2​, at the 0.05 significance level.

SOLUTION 4: LIMITATION OF DATA: HERE WE ASSUMED THAT POPULATIONS VARIANCES ARE EQUAL AND PARENT POPULATIONS ARE NORMALLY DISTRIBUTED. IT MIGHT BE POSSIBLE THAT DATA ARE NOT NORMALLY DISTRIBUTED.

SOLUTION 5: IF POPULATIONS ARE NOT NORMALLY DISTRIBUTED THEN WE NEED TO USE WILCOXON RANK SUM TEST.