case study

Case 10.1: Motive Power Company-Part 1

The key issue is to determine if statistical evidence is sufficient to conclude a difference exists between rivets from the two suppliers.

1. Use Excel’s Histogram option to create histograms for the original supplier and the new Supplier’s distribution of diameter measurements. Make sure you label each accordingly.

2. Use Excel’s descriptive statistics feature to provide a table of descriptive statistics.

3. The assumptions are:

a) The populations are normally distributed. Based on the histograms in part 1, this appears to be the case.

b) The populations have equal variances. The sample variance of the new supplier is about 3.5 time that of the original supplier. This assumption is likely not satisfied which means the students will have to adjust the degrees of freedom. Fortunately, Excel has an option to perform a test with unequal variances.

c) The sample are random. While this appears to be the case for the original supplier, nothing is mentioned in the case about the new supplier. However, the fact the sample distribution is approximately normal is evidence the sample is random.

4. Students will have to pick their own significance level. And justify their choice.

5. Students could perform any of the three types of hypothesis test. The following is Excel output for a significance level of 0.02; Students are encouraged to use a different significance for their analysis. Here the assumption is UNEQUAL VARIANCES WITH ALPHA = 0.02

t-Test: Two-Sample Assuming Unequal Variances	Alpha = 0.02
	Original Supplier	New Supplier
Mean	0.3755	0.3750
Variance	0.000113	0.000393199
Observations	100	100
Hypothesized Mean Difference	0
df	151
t Stat	0.2223
P(T<=t) one-tail	0.4122
t Critical one-tail	2.0716
P(T<=t) two-tail	0.8244
t Critical two-tail	2.3513

Students can also do t-Test for Two Samples Assuming Equal Variances in Excel with the same significance level as they did with the assumption of unequal variances.

t-Test: Two-Sample Assuming Equal Variances, Alpha = 0..05
	Original Supplier	New Supplier
Mean	0.37545	0.37495	-0.0005
Variance	0.000112836	0.000393	0.000280364
Observations	100	100
Pooled Variance	0.000253018
Hypothesized Mean Difference	0
df	198
t Stat	0.222269348
P(T<=t) one-tail	0.412166656
t Critical one-tail	1.285841842
P(T<=t) two-tail	0.824333311
t Critical two-tail	1.652585784

6. Students should present a well-organized report containing the output presented above. Notice there is not a significant difference for either the one or two tailed test. Students should make the appropriate conclusions by comparing the critical values against the t-stat – one or two tailed test, or whether assuming equal or unequal variances.