Question

A randomised block experiment produced the data listed below.

Treatment Block 2 3 4 6 8 2 3 6 6

a) Conduct F-tests at the 5% significance level to find out whether

(i) the treatment means differ;

(ii) the block means differ.

2. (b) Conduct a Friedman test at the 5% significance level to determine whether the treatment medians (central locations) differ.

3. (c) Are the required conditions of the Friedman test valid this time?

Answer 1

Excel add-in (Mega Stat) procedure: Step 1: In EXCEL, Select Add-Ins> Mega Stat> Analysis of Variance. Step 2: Choose Randomized Blocks ANOVA. Step 3: In Input Range, enter the column of AI:E4. Step 4: In Post-Hoc Analysis, Choose When p<0.05. Step 5: Click OK. EXCEL (Mega Stat) output:
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Randomized blocks ANOVA Mean 7.000 5.333 4.667 5.333 n Std. Dev 3 1.000 T1 3 0.577 T2 3 0.577 T3 3 1.155 T4 4.750 6.000 6.000 5.583 4 0.957 B1 4 1.414 B2 4 0.816 B 3 12 1.165 Total ANOVA table Source df 3 2.972 2 2.083 6 0.306 MS Fp-value 9.73 0101 6.82 0285 Treatments 8.92 4.17 1.83 14.92 Blocks Error Total G). Since pvalue0.0101) <a 0.05), then the treatment means differ (ii). Since palue0.0285) a0.05), then the block means differ EXCEL Mega Stat in Add-Ins) procedure

Step 1: Choose Mega Stat > Nonparametric Tests> Friedman Test. Step 2: In Input range, enter the values as Al:E4. Step 3: Check the Correct for ties. Step 4: Click OK. EXCEL Mega Stat in Add-Ins) output:

Friedman Test Sum of Ranks Avg. Rank 12.00 4.00 T1 7.00 2.33 T2 00 1.33 T3 7.00 2.33 T4 30.00 2.50 Total 7.333 chi-square (corrected for ties) 3 d.f. 0620 p-value multiple comparison values for avg. ranks 2.78 (.05) 3.31 (.01) Since p value0.0620)0.05), then the treatment medians do not differ. (d). Yes, the required conditions of the Friedman test are valid this time.