Question

A survey looked at the value of recreational sports on college campuses. One of the questions asked each student to rate the importance of recreational sports to college satisfaction and success. Responses were on a 10-point scale with 1 indicating total lack of importance and 10 indicating very high importance. The following table summarizes these results: Class n Mean score Freshman Sophomore Junior Senior 727 535 592 439 7.6 7.6 7.5 7.3 (a) To compare the mean scores across classes, what are the degrees of freedom for the ANOVA F statistic? numerator df denominator df (b) The MSG = 11.706. If sp = 2.14, what is the F statistic? (Round your answer to two decimal places.) F = (c) Give an approximate (from Table E) or exact (from software) P-value. (Enter your answer to four decimal places.)

Answer 1

a)

Numerator d.f. = 4 – 1 = 3

Denominator d.f. = 727 + 535 + 592 + 439 – 4 = 2289

b)

The F statistic is:

Recall that,

= 11.706 / (2.14)^2

= 2.556118

Therefore,

(Rounded to two decimal places.)

c)

The P-value by using Excel:

Fdist(f, df num, df denom)

Fdist(2.56, 3, 2289)

P(F > 2.56) = 0.0534

This suggests that we should consider the fact that at least one of the classes scored differently. Mainly it looks like the seniors have a different mean. Notice that the difference in means is tiny, and the pooled standard deviation suggest overlap between the data values. But the sample sizes are very large, and thus any small deviation from the perfect (no difference in means) can be detected. However, we have a feeling something is wrong here.
We calculated x-bar (the mean of all the data to be equal 7.51, and calculating MSG we got 9.810, which produces a p-value of 0.0978.