In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations).The following information is provided.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)

If we want to determine whether or not the means of the five populations are equal, the p-value is
A) greater than .10.
B) between .025 to .05.
C) between .01 to .025.
D) less than .01.

Question

In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations).The following information is provided. ​
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
​
If we want to determine whether or not the means of the five populations are equal, the p-value is
A) greater than .10.
B) between .025 to .05.
C) between .01 to .025.
D) less than .01.

Quizplus · Accepted Answer

To determine whether or not the means of the five populations are equal, we need to conduct an ANOVA test. The p-value is the probability of obtaining a test statistic as extreme or more extreme than the one observed, assuming the null hypothesis is true. In this case, the null hypothesis is that the means of the five populations are equal.

The formula for the F-test statistic is:

F = (SSTR / (k - 1)) / (SSE / (n - k)),

where SSTR is the sum of squares due to treatments, SSE is the sum of squares due to error, k is the number of treatments, and n is the total number of observations.

From the given information, we have SSTR = 200, SST = 800, k = 5, and n = 65. We can calculate SSE as:

SSE = SST - SSTR = 800 - 200 = 600.

Plugging these values into the formula for the F-test statistic, we get:

F = (200 / 4) / (600 / 60) = 5.

Using a significance level of α = 0.05, we can look up the critical F-value from an F-distribution table with 4 and 60 degrees of freedom (df). The critical F-value is 2.48.

Since our calculated F-value of 5 is greater than the critical F-value of 2.48, we reject the null hypothesis and conclude that there is strong evidence that the means of the five populations are not equal. The p-value is less than 0.01, which means that the probability of observing a test statistic as extreme or more extreme than the one observed, assuming the null hypothesis is true, is less than 0.01. Therefore, the best choice is D.

In a Completely Randomized Experimental Design Involving Five Treatments, 13