A survey looked at the value of recreational sports on college campuses. One of

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:

(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 = 11.906. 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.)

(b) The MSG = 11.906. 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.)

Explanation / Answer

a) From the available data, the following conclusions can be drawn:

Number of samples = 4

Total sample size = 729 + 536 + 592 + 438 = 2295

Numerator degrees of freedom is given by (Number of samples - 1) = 4 - 1 = 3

Denominator degrees of freedom is given by (Total sample size - Number of samples) = 2295 - 4 - 2291

b) The mean square variability between the sample means is given to be11.906

F ratio = Mean square variability between the sample means or MSG/Mean square errors or Mean square variability within the sample means or MSE

Sp = Root mean square error or Root MSE = 2.14

MSE = 2.14 * 2.14 = 4.5796

F = 11.906/4.5796 = 2.59979037471

Rounding your answer to two decimal places, F = 2.60

c) The exact or approximate p-value for F-statistic(2.60) with degrees of freedom (3 and 2291) is given to be 0.0506

Since the p-value is > 0.05 we accept the null-hypothesis; that is, the recreation sports does have an influence on college satisfaction and success.

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