The following data are from a completely randomized design....

The following data are from a completely randomized design. In the following calculations, use $\alpha=.05$. $$\begin{array}{lccc} & \text { Treatment } & \text { Treatment } & \text { Treatment } \\& \text { 1 } & \text { 2 } & \text { 3 } \\& & & \\& 63 & 82 & 69 \\& 47 & 72 & 54 \\& 54 & 88 & 61 \\& 40 & 66 & 48 \\ \bar{x}_{j} & 51 & 77 & 58 \\ \mathrm{s}_{j}^{2} & 96.67 & 97.34 & 81.99\end{array}$$ a. Use analysis of variance to test for a significant difference among the means of the three treatments. b. Use Fisher's LSD procedure to determine which means are different.

The following data are from a completely randomized design. In the following calculations, use $\alpha=.05$.

$$\begin{array}{lccc} & \text { Treatment } & \text { Treatment } & \text { Treatment } \\& \text { 1 } & \text { 2 } & \text { 3 } \\& & & \\& 63 & 82 & 69 \\& 47 & 72 & 54 \\& 54 & 88 & 61 \\& 40 & 66 & 48 \\
\bar{x}_{j} & 51 & 77 & 58 \\ \mathrm{s}_{j}^{2} & 96.67 & 97.34 & 81.99\end{array}$$
a. Use analysis of variance to test for a significant difference among the means of the three treatments.
b. Use Fisher's LSD procedure to determine which means are different.

Key Concepts

Completely Randomized Design

A completely randomized design is an experimental setup where subjects or experimental units are randomly assigned to different treatment groups. This approach minimizes bias and helps ensure that differences between groups are due to the treatments rather than extraneous variables, providing a sound basis for statistical inference.

Analysis of Variance (ANOVA)

ANOVA is a statistical technique used to compare the means of three or more groups to determine if at least one of the group means is significantly different from the others. It works by partitioning the total variation observed in the data into components attributed to variability between groups and variability within groups, and then comparing these using an F-test.

Fisher's Least Significant Difference (LSD) Procedure

Fisher's LSD is a post-hoc multiple comparison method applied after a significant overall ANOVA result. It allows for pairwise comparisons between treatment means to identify which specific groups differ from each other, using the pooled variance estimate from the ANOVA and controlling for the Type I error rate to a limited extent.

Transcript

00:01 Well, the first thing i want to point out is we have the three groups, the a, b, and c, i believe, are 1, 2, and 3.

00:06 And we were given their means as in, let me quick, 51, 77, and 58.

00:13 51, 71, 77, and 58.

00:18 These were the x bars.

00:20 And then the variances that they gave for each of these were actually wrong.

00:25 These are 72 .5, 73, and 61 .4 .4 .4 .1 .1 .1 .7.

00:30 So these were not right in the book.

00:36 So they are off.

00:38 I put the data into my analysis and i did find when i assume that all three of these means are equal, the anoba test for this comes out to be a large value.

00:54 And let me quick give that to you with, and we're supposed to use the technique to show.

01:01 Of whether they're different or not, but let me just quick tell you what the f statistic was for this.

01:07 And i have it written down, but i don't see it.

01:10 So we'll just click get it.

01:12 And we find that the f statistic for this was a 13 .116, and the p value is 0 .009.

01:22 And this is smaller than that 0 .05.

01:25 So we definitely have evidence to reject the null.

01:27 And the alternative is that they're not all equal.

01:32 And so we would have strong evidence to say that they're not all equal.

01:37 Now we're going to use the fisher's test, lsd test, to see who is different from whom.

01:45 And so we'll first of all deal with the a equaling b.

01:52 And so our test, and alternately, that a does not equal.

01:58 Be and we'll have our test statistic be a t value with 12 degrees of freedom because we have 15 elements less the three factors and so we take the difference between those two means so 51 and 77 those are definitely going to be different and then divided by the square root of and we want that mean square error and that ends up being 69 times one over the first sample size plus one over the second and sample size.

02:30 And so that test statistic, and we can do this, we're doing a two -tail test, so we can do the absolute value or deal with this as a non -absorute value.

02:40 But we have, let's see, 51 minus the 77 and then divided by that square root of 69 times two -fifths.

02:51 And that test statistic comes out to be negative 4 .949.

02:56 And the likelihood of getting that test statistic or, smaller since we have a negative and then we'll be doubling it this is going to be approximately zero and so this is definitely going to be smaller than the 0 .05 significance level so we have evidence to reject the null so we can see that a and b are going to be different now when we do that same thing for the next level and let's go to red if we want to test a and c as opposed to the alternative will be not equal to...

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