In: Statistics and Probability
The following data are from a completely randomized design.
| Treatment | Treatment | Treatment | |
| A | B | C | |
| 32 | 44 | 34 | |
| 30 | 43 | 37 | |
| 30 | 44 | 36 | |
| 26 | 46 | 37 | |
| 32 | 48 | 41 | |
| Sample mean | 30 | 45 | 37 |
| Sample variance | 6 | 4 | 6.5 |
a. At the
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 -value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 6
What is your conclusion?
- Select your answer -Conclude that not all treatment means are equalDo not reject the assumption that the treatment means are equalItem 7
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 .
| Difference | Absolute Value | Conclusion |
| - Select your answer -Significant differenceNo significant differenceItem 10 | ||
| - Select your answer -Significant differenceNo significant differenceItem 12 | ||
| - Select your answer -Significant differenceNo significant differenceItem 14 |
c. Use Fisher's LSD procedure to develop a confidence interval estimate of the difference between the means of treatments A and B (to 2 decimals). Since treatment B has the larger mean, the confidence interval for the difference between the means of treatments A and B ( ) should be reported with negative values.
( , )


from above
a)
| sum of sq;treatment= | 676.00 |
| sum of sq; error= | 66.00 |
| mean sq;treatment= | 338.00 |
| mean square; error= | 4.40 |
| test statistic = | 76.82 |
| p value is less than 0.01 | |
-Conclude that not all treatment means are equal
b)
| Fisher's (LSD) for group i and j =(tN-k)*(sp*√(1/ni+1/nj) = | 2.58 | ||||
| Difference | Absolute Value | Conclusion |
| x2-x1 | 15.00 | significant difference |
| x3-x1 | 7.00 | significant difference |
| x3-x2 | 8.00 | significant difference |
c)
| Lower bound | Upper bound | ||||
| (xi-xj)-ME | (xi-xj)+ME | ||||
| μA-μB | -17.58 | -12.42 |