- A research wishes to try three different techniques to lower stress levels of law enforcement personnel diagnosed with high blood pressure. The subjects are randomly assigned to three groups; the first group takes the medication, the second group exercises, and the third group follows a strict diet. After four weeks, the reduction in each person’s blood pressure is record. At a 95% significance level, test the claim that there is no difference among the means. The data are shown.
What is the critical value?
Compute the test value.
Find the between group variance.
Find the within group variance.
Find the F test value.
Make the decision. [Remember if F is greater than the CV reject the null. If F is less than the CV accept the null].
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`Number , of df=n-1=15-1=14`
`t > t_(0.05)(14)`
This highly significant.
therefor assumption is wrong.
First, in order to find the critical value, one must know the alpha level. This is usally 0.05. Assuming alpha is 0.05, first you need to subtract alpha from 1 (1 - 0.05 = 0.95). If this is a two-tailed test, you then need to divide this by 2 (0.95/2 = 0.475).
Look at a z table to find the critical value. Once you have found 0.475, the number at the top of the column is 0.06, the number to the left is 1.9. Add these numbers together to get 1.96. This is your critical value. Again, assuming this is a two-tailed test, you would need to place +/- in front of your critical value.
In order to find the variance, one must know/find the degrees of freedom. The df for this sample is 14 (15-1); however...
You have noted three variables (medication, exercise, and diet) here. One would need data from each to find the "between group variance", "within group variance", and the "f test value." Therefore, you would have three sets of data.
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