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
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.
The appropriate test statistic and p-value for the hypothesis that Feed B gives rise to larger weight gains are:
t = -3.26905; p-value = .0566
The appropriate test statistic and p-value for the hypothesis test are as follows:
t = -2.4048; p-value = .0431
To compare the weight gains of two kinds of feed, the pigs were split into two groups, and each group received one feed. The weight gains for each group were recorded. In this case, we have Feed A and Feed B.
For the SAS output:
1. The mean weight gain for Feed A is 7.45 kg, with a standard deviation of 1.33529 kg and a standard error of 0.54513 kg. There are 6 observations in this group.
2. The mean weight gain for Feed B is 10.88571 kg, with a standard deviation of 3.49401 kg and a standard error of 1.32061 kg. There are 7 observations in this group.
Next, we need to test the hypothesis that Feed B gives rise to larger weight gains. We assume the variances are unequal. SAS provides the following test statistics and p-values:
- For the hypothesis that the variances are unequal, the test statistic is -2.4048 and the p-value is 0.0431.
- For the hypothesis that the variances are equal, the test statistic is -2.2596 and the p-value is 0.0451.
The appropriate test statistic and p-value for testing if Feed B gives rise to larger weight gains are:
t = -2.4048; p-value = 0.0431
For the JMP output:
The estimated difference in weight gains between Feed B and Feed A is -3.26905 kg, with a standard error of 1.53464 kg. The degrees of freedom for the t-test are 11. The p-value for the test assuming equal variances is 0.0566.
The appropriate p-value for testing if Feed B gives rise to larger weight gains is: 0.0566.
Therefore, the correct answer is:
The appropriate test statistic and p-value are:
t = -2.4048; p-value = 0.0431