Question

A systems specialist has studied the workflow of clerks doing the same inventory work. Based on this study, she designed a new workflow layout for the inventory system. To compare average production for the old and new methods, six clerks were randomly selected for the study. The average production rate for each clerk was recorded before and after the new system was introduced. The results are shown below. Test the claim that the new system is different in the mean number of items processed per shift. Use α = .05.

Clerk Joe Jon Joy Jen Jan Job

Old 123 114 112 82 127 122

New 116 108 93 88 119 111

a. State the null and alternative hypotheses (2 point)
b. Write the formula you will use in the space below: (1 point)
c. Identify all variables and corresponding values from the formula. (2 points)
d. Calculate the test statistic. (2 points) e. What is the critical value? (1 point) _________________
f. Interpret your result within the context of the problem. (2 points)

Answer 1

a)

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: i.e. the mean number of items processed per shift under the old and new systems are the same

Ha: i.e. the mean number of items processed per shift under the old and new systems are different.

This corresponds to a two-tailed test, for which a t-test for two paired samples is used.

b)

The t-statistic is computed as shown in the following formula:

c)

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	123	116	7
	114	108	6
	112	93	19
	82	88	-6
	127	119	8
	122	111	11
Average	113.333	105.833	7.5
St. Dev.	16.367	12.576	8.118
n	6	6	6

For the score differences we have

d)

Test Statistics

The t-statistic is computed as shown in the following formula:

e)

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=5.

Hence, it is found that the critical value for this two-tailed test is tc​=2.571, for α=0.05 and df=5.

f)

Since it is observed that ∣t∣=2.263≤tc​=2.571, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0731, and since p=0.0731≥0.05, it is concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level. Hence the mean number of items processed per shift under the old and new systems are the same.

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