Question

Samples of pages were randomly selected from three different novels. The Flesch Reading Ease scores were...

Samples of pages were randomly selected from three different novels. The Flesch Reading Ease scores were obtained from each​ page, and the​ TI-83/84 Plus calculator results from analysis of variance are given below. Use a 0.05 significance level to test the claim that the three books have the same mean Flesch Reading Ease score.

One-way ANOVA

F=2.6286011383

p=0.0867985984

Factor

df = 2

SS= 386.706818

down arrow↓ MS = 193.353409

One-way ANOVA

up arrow

↑ MS = 193.353409

Error

df=34

SS=2500.95605

MS=73.5575309

Sxp= 8.57656871

Click the icon to view the​ TI-83/84 Plus calculator results.

What is the conclusion for this hypothesis​ test?

A.

Fail to reject H0. There is sufficient evidence to warrant the rejection of the claim that the three books have the same mean Flesch Reading Ease score.

B.

Reject H0. There is insufficient evidence to warrant the rejection of the claim that the three books have the same mean Flesch Reading Ease score.

C.

Reject H0. There is sufficient evidence to warrant the rejection of the claim that the three books have the same mean Flesch Reading Ease score.

D.

Fail to reject H0. There is insufficient evidence to warrant rejection of the claim that the three books have the same mean Flesch Reading Ease score.

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Answer #1

Here p value is p=0.0867985984 which is greater than the level of significance i.e. 0.05.

Hence we fail to reject H0 at 5% level of significance and conclude that three books have the same mean.

Hence option D. is correct.

D.

Fail to reject H0. There is insufficient evidence to warrant rejection of the claim that the three books have the same mean Flesch Reading Ease score.

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