In a completely randomized experimental design 11...

In a completely randomized experimental design, 11 experimental units were used for each of the 4 treatments. Part of the ANOVA table is shown below. Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F Between Treatments | 1500 | ? | ? | ? Within Treatments | ? | ? | ? | ? Total | 5500 Using a = .05, test to see if there is a significant difference among the means of the three populations. 1. Formulate the hypotheses and claim. 2. Select the significance level. 3. Select the test statistic and calculate its value. 4. Identify the critical/p value(s) for the test statistic and state the decision rule. 5. Compare calculated and critical values and reach a conclusion about the null hypothesis.

Transcript

00:01 It is given that in a completely randomized design crd, the number of treatments k is given as 4 and from these 4 treatments, from each treatments 11 experimental units are taken.

00:15 Thus the sample size n equal to 11 into 4 which is 44.

00:22 Thus the total sample size is 44 and the incomplete anova table is given us sources of variation, sum of squares, degrees of freedom, mean sum of squares and f statistics.

00:40 The sources of variation are between groups, within groups and total.

00:48 The sum of squares given are for between sum of square it is 1 ,500 and the total sum of is given as 5 ,500.

01:00 Now the within sum of square can be obtained by subacting between sum of squares from the total sum of square, that is subtracting 1 ,500 from 5 ,500, which is 4 ,000.

01:17 Now the degrees of freedom for between groups is k minus 1 which is 4 minus 1 equal to 3.

01:25 And for within groups the degrees of freedom is n minus k which is 44 minus 4 that is 40 now for total the degrees of freedom is n minus 1 which is 44 minus 1 equal to 43 now the mean sum of squares are calculated us by dividing the respective sum of squares by their respective degrees of freedom by their respective degrees of freedom now for between sum of square dividing 1000 by 500 by 3 gives 500.

02:06 Now for within sum of squares 4 ,000 is divided by 40 which gives 100.

02:15 The f statistics is calculated as subacting, sorry dividing mean sum of squares for between groups by mean sum of squares for within groups that is dividing 500 by 100 which gives 5 now there is a total anova table for question 1 the nala hypothesis h0 is such that mean of the first treatment mu 1 equal to mu 2 equal to mu 3 equal to 4 and the alternative hypothesis h .a.

02:59 Is such that at least one treatment has different mean from others.

03:12 This is the null and alternative hypothesis for the given data.

03:19 And for question 2, the alpha of the significance level is already given as alpha equal to 0.

03:27 So 5.

03:29 Thus the significance level is 0 .05...

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