Question

10-39. An article in Polymer Degradation and Stability (2006, Vol. 91) presented data from a nine-year aging study on S537 foam. Foam samples were compressed to 50% of their original thickness and stored at different temperatures for nine years. At the start of the experiment as well as during each year, sample thickness was measured, and the thicknesses of the eight sam- ples at each storage condition were recorded. The data for two storage conditions follow. 50°C 0.047, 0.060, 0.061, 0.064, 0.080, 0.090, 0.118, 0.165, 0.183 60°C 0.062, 0.105, 0.118, 0.137, 0.153, 0.197, 0.210, 0.250, 0.375

Is there evidence to support the claim that the mean compression increases with the temperature at the storage condition? Use ??=0.05.

a) In Minitab, calculate “Descriptive Statistics” for the two samples. Copy and paste the output here.

b) State the hypotheses.

c) List the assumptions. Check the assumptions are satisfied. In Minitab, make two normal probability plots, one for each sample. Copy and paste the graphs into your report. Do these plots support of the assumption of normality? Why?

d) Calculate the test statistic.

e) Find the p-value.

f) Make a decision.

g) State the conclusion in terms of the problem.

h) In Minitab, perform the appropriate test. Copy and paste the output into your report. In your output, circle the hypotheses, test statistic, and p-value. They should match your answers from above.

i) Calculate a 95% one-sided CI for ??1 ? ??2.

j) Provide an interpretation of the CI.

k) Using the CI, make a decision regarding ??0. Explain how the decision was made.

Answer 1

a)

Descriptive Statistics: 60C, 50C

Variable N N*    Mean SE Mean   StDev Minimum      Q1 Median      Q3 Maximum
60C       9   0 0.1786   0.0312 0.0936   0.0620 0.1115 0.1530 0.2300   0.3750

b)
Ho: mu1 - mu2 = 0
Ha: mu1 - mu2 < 0

c)

check yourself

d)

TS = -2.33
e)
p-value = 0.016
f)
we reject the null as p-value < 0.05
g)
we conclude that there is significant evidence that the mean compression increases with the temperature at the storage condition
h)
Two-Sample T-Test and CI: 50C, 60C

Two-sample T for 50C vs 60C

     N    Mean   StDev SE Mean
50C 9 0.0964 0.0488    0.016
60C 9 0.1786 0.0936    0.031

Difference = ? (50C) - ? (60C)
Estimate for difference: -0.0821
95% upper bound for difference: -0.0207
T-Test of difference = 0 (vs <): T-Value = -2.33 P-Value = 0.016 DF = 16
Both use Pooled StDev = 0.0746

i)
95% CI for difference: (-0.1567, -0.0075)
j) we are 95% confident that the true population difference lies in this interval

k) since every data is below 0
we reject the null