Consider the accompanying data on breaking load (kg/25 mm width) for various fabrics in both an unabraded condition and an abraded condition. Use the paired t test to test H0: μD = 0 versus Ha: μD > 0 at significance level 0.01. (Use μD = μU − μA.)
Fabric | ||||||||
---|---|---|---|---|---|---|---|---|
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
U | 36.3 | 55.0 | 51.2 | 38.5 | 43.2 | 48.8 | 25.6 | 49.5 |
A | 28.5 | 20.0 | 46.0 | 34.0 | 36.0 | 52.5 | 26.5 | 46.5 |
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)
t | = | |
P-value | = |
State the conclusion in the problem context.
Reject H0. The data suggests a significant mean difference in breaking load for the two fabric load conditions.Reject H0. The data does not suggest a significant mean difference in breaking load for the two fabric load conditions. Fail to reject H0. The data suggests a significant mean difference in breaking load for the two fabric load conditions.Fail to reject H0. The data does not suggest a significant mean difference in breaking load for the two fabric load conditions.
U | A | dbar |
36.3 | 28.5 | 7.8 |
55 | 20 | 35 |
51.2 | 46 | 5.2 |
38.5 | 34 | 4.5 |
43.2 | 36 | 7.2 |
48.8 | 52.5 | -3.7 |
25.6 | 26.5 | -0.9 |
49.5 | 46.5 | 3 |
dbar = μ(before) - μ(after) = 7.2625
s(dbar)= 11.8714
SE = s(dbar)/sqrt(n)
= 11.8714./sqrt(8)
= 4.1972
Test Statisitcs,
t = dbar/SE
= 7.2625/4.1972
=1.73
p value = 0.1273
Fail to reject H0. The data does not suggest a significant mean
difference in breaking load for the two fabric load
conditions.
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