Question

A recent college graduate is in the process of deciding which one of three graduate schools he should apply to. He decides to judge the quality of the schools on the basis of the Graduate Management Admission Test (GMAT) scores of those who are accepted into the school. A random sample of six students from each school was taken. Basic statistics and an ANOVA table are provided below.
Source DF SS MS F P Variable N Mean StDev

Factor 2 47511 23756 8.61 0.003 School 1 6 646.7 47.6

Error 15 41400 2760 School 2 6 606.7 68.9

Total 17 88911 School 3 6 523.3 35.6

Assuming all assumptions hold, can he infer at the 10% significance level that the GMAT scores differ among the three schools?
Hypotheses:

Test Statistic P-value statement and p-value

Conclusions and Interpretation (i.e. make a decision and answer the question being posed in one sentence).

Given the F-critical value from the Bonferroni method is + 2.684 and that the pooled standard deviation is 52.54, which of the pairwise differences is significant? Calculate three test statistics to help you answer this question. No hypotheses or conclusions necessary.

Answer 1

Answer:

For this case after conduct the ANOVA procedure they got an statistic of:

$F = 8.61$

With a p value of :

$p_v =P(F_{2,15} > 8.61) = 0.003$

Since the p value is lower than the significance level $\alpha=0.1$

We can reject the null hypothesis that the means are equal at the significance level provided.

$t = \frac{\bar X_i -\bar X_j}{s_p \sqrt{\frac{1}{n_i} +\frac{1}{n_j}}}$

For school 1 and 2 we have:

$t = \frac{646.7 -606.7}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 1.318$

For school 1 and 3

$t = \frac{646.7 -523.3}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 4.068$

For school 2 and 3 we have:

$t = \frac{606.7 -523.3}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 2.749$

So as we can see we have significant difference between the means of school 1 and 3, and school 2 and 3

Step-by-step explanation:

Previous concetps

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

Solution to the problem

The hypothesis for this case are:

Null hypothesis: $\mu_{1}=\mu_{2}=\mu_{3}$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=1,2,3$

If we assume that we have $p$ groups and on each group from $j=1,\dots,p$ we have $n_j$ individuals on each group we can define the following formulas of variation:  

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$  

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$  

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$  

And we have this property  

$SST=SS_{between}+SS_{within}$  

For this case after conduct the ANOVA procedure they got an statistic of:

$F = 8.61$

With a p value of :

$p_v =P(F_{2,15} > 8.61) = 0.003$

Since the p value is lower than the significance level $\alpha=0.1$

We can reject the null hypothesis that the means are equal at the significance level provided.

For the other part we can calculate the 3 statistics with the following formula:

$t = \frac{\bar X_i -\bar X_j}{s_p \sqrt{\frac{1}{n_i} +\frac{1}{n_j}}}$

For school 1 and 2 we have:

$t = \frac{646.7 -606.7}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 1.318$

For school 1 and 3

$t = \frac{646.7 -523.3}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 4.068$

For school 2 and 3 we have:

$t = \frac{606.7 -523.3}{52.54 \sqrt{\frac{1}{6} +\frac{1}{6}}}= 2.749$

So as we can see we have significant difference between the means of school 1 and 3, and school 2 and 3