Question

# Item Group 1 Group 2 Group 3 1 14 17 17 2 13 16 14 3...

 Item Group 1 Group 2 Group 3 1 14 17 17 2 13 16 14 3 12 16 15 4 15 18 16 5 16 14 6 16

a. Conduct a one-way analysis of variance on the data assuming the populations have equal variances and the populations are normally distributed. Use alpha = 0.05.

b. If warranted, use the Tukey-Kramer procedure to determine which populations have different means. Use an alpha level of 0.05.

 Groups Count Sum Average Variance Group 1 5 70 14 2.5 Group 2 4 67 16.75 0.916667 Group 3 6 92 15.33333 1.466667

a) One Way ANOVA:

and (Atleast one is not equal)

Significant value= 0.05

The test statistic:

Grand mean:

P-value: 0.025853

The test statistic is significant and rejects H0. There is sufficient evidence to support the claim that there is statistical difference between sample means.

b) Tukey's mean difference:

To be non significant:

Mean difference between 1 and 2: |14-16.75|= 2.75

|M1-M2| > T is significant.

Mean difference between 1 and 3: |M1-M3|= 1.333

|M1-M2| < T is not significant.

Mena difference between 2 and 3: |M2-M3|= 1.41667

|M2-M3| < T is not significant.

Group1 and 2 are significantly different.

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