Statistics

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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. 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 p-value are:

t = -2.4048; p-value = .0431

t =-2.4048; p-value = .0216

t = -2.2596; p-value = .0451

t = -2.2596; p-value = .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
t-Test
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 p-value for the test is:

.0566

.0283

.1132

.1087

2.130

• Statistics - ,

I just did the test using a TI-83 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.