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.
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
1 answer
1 answer