Question

1. Nine samples of PVC pipe of equal wall thickness are tested under three temperature conditions to failure, yielding the results shown below. Research questions: Is mean burst strength affected by temperature level? Is there an “ideal” temperature level? Explain. You will need to enter the data into Minitab.

Burst Strength of PVC Pipes (Pounds Per Square Inch)
Temperature
Hot (70 Degrees C)	Warm (40 Degrees C)	Cool (10 Degrees C)
250	321	358
301	342	375
235	302	328
273	322	363
285	322	355
260	315	336
281	299	341
275	339	354
279	301	342

• At the 0.05 level of significance, determine if there is a difference mean burst strength by temperature level. State your hypotheses and show all 7 steps clearly. (14 points)
• Give and interpret the p-value. (3 points)
• Should Tukey pairwise comparisons be conducted? Why or why not? (3 points)
• If appropriate, use Minitab to produce Tukey pairwise comparison. Write a few sentences with your conclusions from those comparisons. (4 points)
• Use Levene’s test to determine if the assumption of homogeneity of variances is valid. Give the hypotheses, test statistic, p-value, decision and conclusion. Use the 0.05 level of significance. (8 points)
• Verify with Minitab by attaching or including relevant output. (6 points)

Answer 1

At the 0.05 level of significance, determine if there is a difference mean burst strength by temperature level. State your hypotheses and show all 7 steps clearly. (14 points)

Ho: µ1 = µ2= .. = µ3

Ha: at least one mean is different

Analysis of Variance

Source DF Adj SS   Adj MS F-Value P-Value
temp     2   28580 14290.1    50.18    0.000
Error   24    6834    284.8
Total   26   35415

TS = 50.18

p-value = 0.000 < alpha

hence we reject the null hypothesis

we conclude that there is significant difference in means

Give and interpret the p-value. (3 points)

p-value = 0.000

it means that if there is no difference in means, then probability of observing extreme such sample is 0.000

Should Tukey pairwise comparisons be conducted? Why or why not? (3 points)

yes, since we reject the null hypothesis, we can use post-hoc tests

If appropriate, use Minitab to produce Tukey pairwise comparison. Write a few sentences with your conclusions from those comparisons. (4 points)

Tukey Simultaneous Tests for Differences of Means

Difference   Difference       SE of                             Adjusted
of Levels      of Means Difference       95% CI       T-Value   P-Value
Hot - Cool       -79.22        7.95 (-99.08, -59.37)    -9.96     0.000
Warm - Cool      -32.11        7.95 (-51.97, -12.25)    -4.04     0.001
Warm - Hot        47.11        7.95 ( 27.25, 66.97)     5.92     0.000

p-value for each comparison is less than alpha

hence there is significant difference between each pair