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[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?


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.



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

 

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 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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