Question

As part of its freshman orientation process, a college gives a math placement exam to incoming freshmen. The math department is interested in whether there is a statistically significant difference in the average exam score for students in different programs, at a level of α=0.01. The exam scores for random samples of science, engineering, humanities, and business majors are shown in the following table. The math department has confirmed that the samples were randomly selected and independent, that the populations are normally distributed, and that the population variances of scores are equal. Use Excel to calculate the degrees of freedom, critical value, and test statistic for the data set, rounding the critical value and test statistic to two decimal places.

Science	Engineering	Humanities	Business
89	83	86	78
89	85	82	78
90	86	86	82
88	82	85	81
83	82	82	79
82	87	88	87
85	83	85	85
83	85	87	83
83	83	86	82
87	87	86	79
87	86	81	76
81	87	85	78
87	90	84	79
83	85	85	74
85	86	86	88
88	81	86	83
85	88	79	82
85	87	82	81
89	88	82	78
89	91	79	78

Numerator degrees of freedom = ?

denominator degrees of freedom =?

critical value = ?

test statistic = ?

Answer 1

using minitab we have the following output

One-way ANOVA: Science, Engineering, Humanities, Business

Method

Null hypothesis All means are equal
Alternative hypothesis At least one mean is different
Significance level α = 0.01

Equal variances were assumed for the analysis.

Factor Information

Factor Levels Values
Factor 4 Science, Engineering, Humanities, Business

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
Factor 3 361.5 120.513 14.15 0.000
Error 76 647.3 8.518
Total 79 1008.9

Model Summary

S R-sq R-sq(adj) R-sq(pred)
2.91852 35.84% 33.30% 28.90%

Means

Factor N Mean StDev 99% CI
Science 20 85.900 2.751 (84.176, 87.624)
Engineering 20 85.600 2.683 (83.876, 87.324)
Humanities 20 84.100 2.594 (82.376, 85.824)
Business 20 80.550 3.546 (78.826, 82.274)

Pooled StDev = 2.91852

Numerator degrees of freedom = 3

denominator degrees of freedom =76

critical value = 4.050

test statistic = 14.15

since the value of test statistic is less then critical value we reject the null hypothesis .