An engineering company has approached you with two problems that require your st
ID: 3361279 • Letter: A
Question
An engineering company has approached you with two problems that require your statistical expertise. The company makes plastics for a variety of products. A key measurement is the failure stress levels of plastic measured in mega pascals (MPa).
The second dataset twosample has the failure stress levels (MPa) of a random sample of 30 ‘Type A’ plastics and a random sample of 30 ‘Type B’ plastics. The column labels are as follows: Type: Type A or Type B Strength: Failure stress levels (MPa)
use Minitab to:
What is the difference likely to be on average?
Type Strength
Type A 11.96
Type A 9.65
Type A 8.25
Type A 10.30
Type A 11.53
Type A 10.16
Type A 11.47
Type A 8.39
Type A 11.25
Type A 9.70
Type A 9.33
Type A 10.78
Type A 10.91
Type A 11.18
Type A 10.16
Type A 9.96
Type A 11.25
Type A 11.10
Type A 9.45
Type A 10.78
Type A 10.07
Type A 9.54
Type A 9.80
Type A 10.06
Type A 11.71
Type A 9.40
Type A 9.27
Type A 8.77
Type A 12.32
Type A 9.51
Type B 8.96
Type B 6.65
Type B 5.25
Type B 7.30
Type B 8.53
Type B 7.16
Type B 8.47
Type B 5.39
Type B 8.25
Type B 6.70
Type B 6.33
Type B 7.78
Type B 7.91
Type B 8.18
Type B 7.16
Type B 6.96
Type B 8.25
Type B 8.10
Type B 6.45
Type B 7.78
Type B 7.07
Type B 6.54
Type B 6.80
Type B 7.06
Type B 8.71
Type B 6.40
Type B 6.27
Type B 5.77
Type B 9.32
Type B 6.51
Explanation / Answer
## By using two sample t-test in minitab
Two-Sample T-Test and CI: Type A, Type B
Two-sample T for Type A vs Type B
N Mean StDev SE Mean
Type A 30 10.27 1.05 0.19
Type B 30 7.27 1.05 0.19
Difference = mu (Type A) - mu (Type B)
Estimate for difference: 3.00000
95% CI for difference: (2.45750, 3.54250)
T-Test of difference = 0 (vs not =): T-Value = 11.07 P-Value = 0.000 DF = 58
Both use Pooled StDev = 1.0496
Null Hypothesis: There is no significant average difference between the failure stress levels of type A and type B of plastics.
Vs
Alternative Hypothesis: There is a significant average difference between the failure stress levels of type A and type B of plastics.
Here Pvalue for the test is 0.000 indicates the rejection of null hypothesis at 5% level of significance with 58 degrees of freedom.
That is the average difference between the failure stress levels of type A and type B of plastics is significant.
The 95% confidence interval indicates that the average difference of population lies in the interval
(2.45750, 3.54250)