4. Suppose a study was conducted on a student\'s test scores under certain testi

Question

4. Suppose a study was conducted on a student's test scores under certain testing environments: listening to normal music, listening to sounds from nature, and no sound at all, and obtained the following results. Music Sound No Sound/Music 4 4 4 6 Since they are trying to prove that the average test scores are relatively different with these scenarios, state the general null and alternative hypothesis. Calculate and clearly identify the degrees of freedom between the groups and the degrees of freedom within the groups (vb, vw). Calculate your critical F score using a significance level of 5 %. Calculate (with an outline of work shown) the: a. b. C. d. a. Sum of square error total (SSt) b. Sum of square error within (SSw) c. Sum of square error between (SSb) e. Calculate and clearly identify the: a. b. c. Mean square error between (MSb) Mean square error within (MSw) Your test statistic for your hypothesis. f. Conclude vour hvpothesis test and give an interoretation of vour results

Explanation / Answer

Ans:

a)H0:Average test scores are same using these 3 scenarios

Ha :Average test scores are different using these 3 scenarios

b)df(between)=3-1=2

df(within)=13-3=10

c)Critical F score=Finv(0.05,2,10)=4.103

d)

SST=301-(57^2/13)=51.077

SSTR=(31^2/5)+(17^2/4)+(9^2/4)-(57^2/13)=34.777

SSE=SST-SSTR=16.3

e)

MS(between)=SSTR/df(between)=51.077/2=17.389

MS(within)=SSE/df(within)=16.3/10=1.63

F=17.389/1.63=10.668

f)As,F=10.668>4.103,we reject null hypothesis.

There is sufficient evidence that average test scores are relitavily different with these 3 scenariors.

Music Sound No sound/music 7 5 2 4 5 4 6 3 1 8 4 2 6 Total= 31 17 9 squares of all observations 49 25 4 16 25 16 36 9 1 64 16 4 36 sum of xi's 57 sum of xi'squares 301 SST= 51.077 SSTR= 34.777 SSE= 16.300 df(between)= 2 df(within)= 10 SS df MS F p-value Between(SSTR) 34.777 2 17.3885 10.668 0.003 within(SSE) 16.3 10 1.63 SST 51.077

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