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 = 300 (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 = 300 (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 analysis of variance (ANOVA). The ANOVA table is usually summarized as follows:

Source         | Degrees of Freedom | Sum of Squares | Mean Square | F-statistic
-------------------------------------------------------------------------------
Treatments     | k-1                | SSTR           | MST         | MST/MSE
Error          | N-k                | SSE            | MSE         |
Total          | N-1                | SST            |             |

where k is the number of treatments, N is the total number of observations, SST is the total sum of squares, SSTR is the sum of squares due to treatments, SSE is the sum of squares due to error (or residuals), MST is the mean square due to treatments (MST=SSTR/(k-1)), and MSE is the mean square due to error (MSE=SSE/(N-k)). The F-statistic is the ratio of MST over MSE, which follows an F-distribution with k-1 and N-k degrees of freedom.

The null hypothesis is that all the population means are equal, and the alternative hypothesis is that at least one population mean is different from the others. The ANOVA table allows us to test this hypothesis by comparing the F-statistic to a critical value from the F-distribution. Alternatively, we can use the p-value approach, which is to calculate the probability of obtaining a F-statistic as extreme as the observed one, assuming that the null hypothesis is true. If the p-value is smaller than a chosen significance level (such as 0.05 or 0.01), we reject the null hypothesis in favor of the alternative hypothesis.

In this problem, we are given that SST = 800 and SSTR = 300. The degrees of freedom for treatments is k-1=5-1=4, and the degrees of freedom for error is N-k=65-5=60. Therefore, the mean square due to treatments is MST=SSTR/4=75, and the mean square due to error is MSE=SSE/60.

The F-statistic is MST/MSE=75/MSE, but we don't know MSE yet. However, we can use the fact that SST=SSTR+SSE to obtain:

SSE = SST - SSTR = 800 - 300 = 500

Therefore, the mean square due to error is MSE=SSE/60=500/60=8.33 (rounded to two decimal places).

Finally, the F-statistic is:

F = MST/MSE = 75/8.33 = 9.00 (rounded to two decimal places)

This F-statistic follows an F-distribution with 4 and 60 degrees of freedom. Using a table or calculator, we can find that the p-value is less than 0.01 (approximately 0.0001). Therefore, we reject the null hypothesis and conclude that at least one population mean is different from the others, with a high level of confidence. The correct answer is D.

In a Completely Randomized Experimental Design Involving Five Treatments, 13