Question

A portion of an ANOVA summary table is shown below.
Source DF SS MS
Between 2   84.11 42.06
Within (error) 15 88.83 5.92
Total 17 172.94

Test at the 5% level of significance to determine whether differences exist among the
population means.

Answer 1

Solution :

Null and alternative hypotheses:

The null and alternative hypotheses would be as follows :

H_{​​​​​​0} : All the population means are equal.

H_{​​​​​​1} : At least one of the population mean is different from others.

Test statistic :

The test statistic is given as follows :

We have, MS_between = 42.06 and MS_{​​​​​within} = 5.92

The value of the test statistic is 7.1047.

P-value :

The p-value for the test is given as follows :

P-value = P(F > value of the test statistic)

P-value = P(F > 7.1047)

P-value = 0.0067

Decision :

Significance level = 5% = 0.05

P-value = 0.0067

(0.0067 < 0.05)

Since, p-value is less than the significance level of 5%, therefore we shall reject the null hypothesis (H_{​​​​​​0}) at 5% significance level.

Conclusion :

At 5% significance level, there is sufficient evidence to conclude that differences exist between the population means.

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