## [answered] A repeated-measures experiment comparing only two treatment

A repeated-measures experiment comparing only two treatments can be evaluated with either a?t?statistic or an ANOVA. As we found with the independent-measures design, the?t?test and the ANOVA produce equivalent conclusions, and the two test statistics are related by the equation?F?=?t2.

The following data are from a repeated-measures study.

 Subject Treatment 1 Treatment 2 Difference 1? 1 9 +8 2? 2 3 +1 3? 2 4 +2 4? 1 7 +6

(a) Use a repeated-measures?t?statistic with???= 0.05 to determine whether the data provide evidence of a significant difference between the two treatments. (Caution:?ANOVA calculations are done with the?X?values, but for?t?you use the difference scores. Round your answers to three decimal places.)

 t-critical =? ? t?=?

Conclusion?

Fail to reject the null hypothesis. There are significant differences between the two treatments.Reject the null hypothesis. There are not significant differences between the two treatments.Reject the null hypothesis. There are significant differences between the two treatments.Fail to reject the null hypothesis. There are not significant differences between the two treatments.

(b) Use a repeated-measures ANOVA with???= 0.05 to evaluate the data. (You should find?F?=?t2.) Fill in the missing values. (Round your answers for?SS,?MS, and?F?to two decimal places.)

 Source? SS df MS F Between Treatments? Within Treatments? Between Subjects? Error? Total?

Fcrit?=??

Conclusion

Reject the null hypothesis. There are significant differences between the two treatments.Fail to reject the null hypothesis. There are not significant differences between the two treatments.Fail to reject the null hypothesis. There are significant differences between the two treatments.Reject the null hypothesis. There are not significant differences between the two treatments.

A repeated-measures experiment comparing only two treatments

can be evaluated with either a t statistic or an ANOVA. As we

found with the independent-measures design, the t test and the

ANOVA produce equivalent conclusions, and the two test

statistics are related by the equation F = t2.

The following data are from a repeated-measures study.

Subjec

t Treatment

1 Treatment

2 Differenc

e 1 1 9 +8 2 2 3 +1 3 2 4 +2 4 1 7 +6 (a) Use a repeated-measures t statistic with ? = 0.05 to

determine whether the data provide evidence of a significant

difference between the two treatments. (Caution: ANOVA

calculations are done with the X values, but for t you use the

t-critical

?

=

t=

Conclusion

Fail to reject the null hypothesis. There are significant differences

between the two treatments.

Reject the null hypothesis. There are not significant differences

between the two treatments.

Reject the null hypothesis. There are significant differences

between the two treatments.

Fail to reject the null hypothesis. There are not significant

differences between the two treatments.

(b) Use a repeated-measures ANOVA with ? = 0.05 to evaluate

the data. (You should find F = t2.) Fill in the missing values.

(Round your answers for SS, MS, and F to two decimal places.) Source SS df MS Between

Treatments

Within Treatments

Between Subjects

Error

Total

What is the critical F value? (Round your answer to two decimal

places.)

Fcrit =

Conclusion

Reject the null hypothesis. There are significant differences

between the two treatments.

Fail to reject the null hypothesis. There are not significant

differences between the two treatments.

Fail to reject the null hypothesis. There are significant differences

between the two treatments.

Reject the null hypothesis. There are not significant differences

between the two treatments. F

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