Statistics
posted by Bill on .
In order to compare two kinds of feed, thirteen pigs are split into two groups, and each group received one feed. The following are the gains in weight (kilograms) after a fixed period of time:
Feed A: 8.0 7.4 5.8 6.2 8.8 9.5
Feed B: 12.0 18.2 8.0 9.6 8.2 9.9 10.3
We wish to test the hypothesis that Feed B gives rise to larger weight gains. Assume the variances to be unequal. The output from SAS is as follows:
Variable: GAIN Weight gain (kg)
FEED N Mean Std Dev Std Error

a 6 7.45000000 1.33529023 0.54512995
b 7 10.88571429 3.49400848 1.32061107
Variances T DF Prob>T

Unequal 2.4048 7.9 0.0431
Equal 2.2596 11.0 0.0451
For H0: Variances are equal, F' = 6.85 DF = (6,5) Prob>F' = 0.0520
The appropriate test statistic and pvalue are:
t = 2.4048; pvalue = .0431
t =2.4048; pvalue = .0216
t = 2.2596; pvalue = .0451
t = 2.2596; pvalue = .0256
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In order to compare two kinds of feed, thirteen pigs are split into two groups, and each group received one feed. The following are the gains in weight (kilograms) after a fixed period of time:
Feed A: 8.0 7.4 5.8 6.2 8.8 9.5
Feed B: 12.0 18.2 8.0 9.6 8.2 9.9 10.3
We wish to test the hypothesis that Feed B gives rise to larger weight gains. The output from JMP is as follows:
Difference
tTest
DF
Prob > t
Estimate
3.26905
** hidden **
11
0.0566
Std Error
1.53464
Lower 95%
6.64677
Upper 95%
0.10868
Assuming equal variances
The pvalue for the test is:
.0566
.0283
.1132
.1087
2.130

I just did the test using a TI83 calculator.
I got a t=2.60 with a p=.026
If you are doing the test... since .026 is smaller than .05, you can reject the null hypothesis. Which means they did gain more weight with B.