Questions: The following data are from a completely randomized design. - Treatment A: 32, 30, 30, 26 - Treatment B: 45, 44, 45, 47 - Treatment C: 32, 35, 34, 35 Sample mean: Treatment A - 32, Treatment B - 49, Treatment C - 39 Sample variance: Treatment A - 30, Treatment B - 46, Treatment C - 35 Standard deviation: Treatment A - 6.00, Treatment B - 4.00, Treatment C - 6.50 a. At the alpha=0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). - Sum of Squares, Treatment - Sum of Squares, Error - Mean Squares, Treatment - Mean Squares, Error Calculate the value of the test statistic (to 2 decimals). The p-value is What is your conclusion? Conclude that not all treatment means are equal b. Calculate the value of Fisher's LSD (to 2 decimals). Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and treatments B and C. Use alpha=0.05. Difference Absolute Value Conclusion - Mean A - Mean B: Significant difference - Mean A - Mean C: Significant difference - Mean B - Mean C: Significant difference

Solution Steps

To solve this problem, we need to perform an ANOVA test to determine if there are significant differences between the means of the three treatments. We will calculate the Sum of Squares for Treatment and Error, then compute the Mean Squares for both. Using these, we will calculate the F-statistic and determine the p-value to test the null hypothesis. If the p-value is less than the significance level, we reject the null hypothesis. For Fisher's LSD, we will calculate the LSD value and compare the differences between treatment means to determine significance.

The Sum of Squares for Treatment is calculated using the formula: \[ SST = n \sum (\bar{x}_i - \bar{x})^2 \] where \( n \) is the number of observations per treatment, \(\bar{x}_i\) is the mean of each treatment, and \(\bar{x}\) is the overall mean. Given:

• \( n = 4 \)
• \(\bar{x}_A = 29.5\), \(\bar{x}_B = 45.25\), \(\bar{x}_C = 34.0\)
• \(\bar{x} = 36.25\)

\[ SST = 4 \left((29.5 - 36.25)^2 + (45.25 - 36.25)^2 + (34.0 - 36.25)^2\right) = 526.5 \]

The Sum of Squares for Error is calculated as: \[ SSE = \sum (x_{ij} - \bar{x}_i)^2 \] where \( x_{ij} \) are the individual observations. The calculated value is: \[ SSE = 29.75 \]

The Mean Square for Treatment (MST) and Mean Square for Error (MSE) are calculated as: \[ MST = \frac{SST}{k-1} = \frac{526.5}{3-1} = 263.25 \] \[ MSE = \frac{SSE}{N-k} = \frac{29.75}{12-3} = 3.3056 \] where \( k = 3 \) is the number of treatments and \( N = 12 \) is the total number of observations.

The F-statistic is calculated as: \[ F = \frac{MST}{MSE} = \frac{263.25}{3.3056} = 79.64 \]

The p-value is calculated based on the F-distribution with degrees of freedom \( (k-1, N-k) = (2, 9) \). The p-value is approximately \( 1.8922 \times 10^{-6} \), which is less than the significance level \(\alpha = 0.05\). Therefore, we reject the null hypothesis and conclude that not all treatment means are equal.

Fisher's LSD is calculated as: \[ LSD = t_{\alpha/2, N-k} \times \sqrt{MSE \times \frac{2}{n}} \] where \( t_{\alpha/2, N-k} = 2.2622 \) is the critical t-value. Thus: \[ LSD = 2.2622 \times \sqrt{3.3056 \times \frac{2}{4}} = 2.9082 \]

Calculate the absolute differences:

• \(|\bar{x}_A - \bar{x}_B| = |29.5 - 45.25| = 15.75\)
• \(|\bar{x}_A - \bar{x}_C| = |29.5 - 34.0| = 4.5\)
• \(|\bar{x}_B - \bar{x}_C| = |45.25 - 34.0| = 11.25\)

Compare each difference to the LSD:

• \(15.75 > 2.9082\), significant difference between A and B.
• \(4.5 > 2.9082\), significant difference between A and C.
• \(11.25 > 2.9082\), significant difference between B and C.

Final Answer