What is the test statistic? What is the p-score?

Question

Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. "Wrinkle recovery angle" measures how well a fabric recovers from wrinkles. Higher is better. Here are data on the wrinkle recovery angle (in degrees) for a random sample of fabric specimens. Assume the populations are approximately normally distributed with unequal variances.

A manufacturer believes that the mean wrinkle recovery angle for Hylite is better. A random sample of 25 Permafresh (group 1) and 20 Hylite (group 2) were measured. Test the claim using a 1% level of significance.

Permafresh (�1)Hylite (�2)
97	141
111	147
153	136
139	143
137	134
155	141
149	137
116	142
169	145
144	141
109	137
130	133
146	135
122	143
132	142
145	134
125	139
137	142
152	137
114	148
132
121
123
133
147

Summary of results:

x̅₁ = 133.52, x̅₂ = 139.85 

s₁ = 16.998, s₂ = 4.356 

Answer 1

Hi Kaitlyn,

First off, let's identify the null and alternative hypotheses:

H₀: mu₁=mu₂, no significant difference in true mean recovery angle

H_A: mu₁ not equal to mu₂; some significant difference in true mean recovery angle

Again, we've got that degrees of freedom problem for unequal sample variances. We can get test statistic relatively easily:

t=(x-bar₁ - x-bar₂)/sqrt[(s₁²/n₁) + (s₂²/n₂)

x-bar₁=133.52

x-bar₂=139.85

s₁₌16.99

s₂=4.356

n₁=25

n₂=20

t=(133.52 - 139.85)/sqrt[(16.99²/25) + (4.356²/20)]

t= -1.79

That's our t-statistic, but we need degrees of freedom for p-value. I located an easier formula than the one we used in our last problem. This is Satterthwaite's Correction:

df= [(n₁-1)(n₂-1)] / [(n₁-1)c₂² + (n₂-1)c₁²]

where:

c₁₌(s₁²/n₁)/[(s₁²/n₁) + (s₂²/n₂)]

c₂=1-c₁

Thus:

c₁= (16.998²/25)/[(16.998²/25) + (4.356²/20)]

c₁= 0.924

c₂= 0.076

Now, we plug those values into the initial Satterthwaite equation; all other variables are the same as when we computed the test statistic

df= [(n₁-1)(n₂-1)] / [(n₁-1)c₂² + (n₂-1)c₁²]

df=[(25-1)(20-1)]/[((25-1)0.076²) + ((20-1)0.924)²]

df=27.87, round down

df=27

Now, look at the t-table row for 27 degrees of freedom and look for the absolute value of our t-test statistic -1.79, abs[-1.79]=1.79. You will see that this falls between the two-tailed p-values 0.05 and 0.10. Thus:

0.05 < p < 0.10

You cannot get any closer than that without statistical software. However, we do know that we cannot reject the null hypothesis since p is well above our 0.01 significance level. Therefore, we fail to reject the null and conclude no significant difference in mean recovery angles. I hope this helps.