Question

Problem 1: Here are data from an experiment studying the effect of age on creativity. Children in different age groupings need to be compared to see if any one group is different than the others when it comes to creativity. Conduct a one way ANOVA to determine the answer (using the steps) Age 4 Age 6 Age 8 Age 10 ave TOTALS 152 598 1332 20 n 10.8 k=4

Answer 1

The following table -

	Group 1	Group 2	Group 3	Group 4
	3	9	9	7
	5	11	12	7
	7	14	9	7
	4	10	8	6
	3	10	9	4
Sum =	22	54	47	31
Average =	4.4	10.8	9.4	6.2
∑i​Xij 2​=	108	598	451	199
St. Dev. =	1.673	1.924	1.517	1.304
SS =	11.2	14.8	9.2	6.8
n =	5	5	5	5

The total sample size is N=20. Therefore, the total degrees of freedom are:

dftotal​ = 20 − 1 = 19

Also, the between-groups degrees of freedom are dfbetween ​= 4 − 1 = 3,

and the within-groups degrees of freedom are:  dfwithin​ =  dftotal ​− dfbetween​ =19 − 3 =16

First, we need to compute the total sum of values and the grand mean. The following is obtained

$\sum_{i,j} X_{ij} = 22+54+47+31 = 154$

Also, the sum of squared values is

$\sum_{i,j} X_{ij}^2 = 108+598+451+199 = 1356$

Based on the above calculations, the total sum of squares is computed as follows

$SS_{total} = \sum_{i,j} X_{ij}^2 - \frac{1}{N} \left(\sum_{i,j} X_{ij}\right)^2 = 1356 - \frac{ 154^2}{ 20} = 170.2$

The within sum of squares is computed as shown in the calculation below:

SS within => SS withingroups = 11.2 + 14.8 + 9.2 + 6.8 = 42

Now that sum of squares are computed, we can proceed with computing the mean sum of squares:

$MS_{between} = \frac{SS_{between}}{df_{between}} = \frac{ 128.2}{ 3} = 42.733$

$MS_{within} = \frac{SS_{within}}{df_{within}} = \frac{ 42}{ 16} = 2.625$

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows:

$F = \frac{MS_{between}}{MS_{within}} = \frac{ 42.733}{ 2.625} = 16.279$

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1 = μ2 = μ3 = μ4

Ha: Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df_1 = 3   and df2 ​ = 3, therefore, the rejection region for this F-test is

(3) Test Statistics

$F = \frac{MS_{between}}{MS_{within}} = \frac{ 42.733}{ 2.625} = 16.279$

(4) Decision about the null hypothesis

Since it is observed that  F  = 16.279 >  Fc​ = 3.239, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is  p=0, and since p=0<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 4 population means are equal, at the α=0.05 significance level.

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