Question

The mean effect of three treatments on fasting blood sugar levels for three samples of 10 patients are shown below. Treatment, Mean Fasting BG level A= 85.6 B = 80.8 C = 76.2 The ANOVA output for this experiment is as follows. (df, SS, MS, F-value, Pr(>F)) Treatment: 2, 441.9, 220.9, 4.340, .0232 Residuals: 27, 1374.6, 50.9 Assuming the level of significance is α=0.05, compare the pairwise differences in the mean blood glucose level for the three treatments using Fisher's Least Significant Difference method. Step 1 of 2 : Compute Fisher's Least Significant Difference for comparing the mean blood glucose level for the three treatments. Round your answer to four decimal places. Step 2 of 2 : Determine which pairs of means are significantly different using the value for Fisher's LSD calculated in the previous step.

Answer 1

Using Fisher's Least Significant Difference method, with the computed LSD value of 6.5435, we determine that there are no significant differences in mean fasting blood glucose levels between treatments A and B or B and C. However, there is a significant difference between treatments A and C.

Step-by-step explanation:

The question is asking us to employ Fisher's Least Significant Difference (LSD) method to determine whether the mean blood glucose levels for three different treatments are different from each other. Given that the significance level α is 0.05, the ANOVA table provides us with the mean square error (Residual MS) of 50.9. Fisher's LSD is calculated by taking the square root of the mean square error divided by the sample size, and then multiplying by the t-value from the t-distribution for the df of the residuals with a two-tailed test at the chosen α level.

Step 1: Compute Fisher's LSD for comparing the means. From the ANOVA table, we have Residual MS = 50.9, and with each sample size being 10, we find the LSD as follows:

LSD = t * sqrt(2 * MSE / n)

Where t is the critical value from the t-distribution table for df=27 at α=0.05 (two-tailed). We need to look up the t value for a two-tailed test with 27 degrees of freedom at the 0.05 significance levels.

Assuming the t-value is approximately 2.052 (which you would get from the t-table), we calculate:

LSD = 2.052 * sqrt(2 * 50.9 / 10) = 2.052 * sqrt(10.18) = 2.052 * 3.189 = 6.5435

We round this to four decimal places as instructed, so the computed LSD is 6.5435.

Step 2: Determine the significant differences between treatment pairs. We compare the absolute differences in means for each treatment pair with the computed LSD:

• Mean difference AB = |85.6 - 80.8| = 4.8
• Mean difference AC = |85.6 - 76.2| = 9.4
• Mean difference BC = |80.8 - 76.2| = 4.6

Since the mean difference for AB (4.8) and BC (4.6) are both less than the LSD (6.5435), we conclude there is no significant difference between treatments A and B or between treatments B and C. However, the difference for AC (9.4) is greater than the LSD, therefore we conclude there is a statistically significant difference between treatments A and C.