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February 1, 2015

February 1, 2015

Posted by **Bill** on Monday, February 18, 2013 at 3:56pm.

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

____

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 -
**JJ**, Monday, February 18, 2013 at 4:48pmI 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.

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