Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

In a study comparing individual performance with group performance, researchers

ID: 3150128 • Letter: I

Question

In a study comparing individual performance with group performance, researchers found that groups consistently outperformed the best of the individuals. The task was a letters-to-numbers problem in which an arithmetic problem was presented substituting a letter in place of each digit. Participants had to determine which letters corresponded to each number. Three-person groups competed with individuals and scores were compared for the groups and the best of the individuals. The dependent variable was the number of trials needed to solve each problem. The following data were obtained in the study.

(a) Based on these results, is there a significant difference between the performance for individuals versus the performance for groups? Use a two-tailed test with = 0.05. (Use 3 decimal places.)

Conclusion ( chose A, B, C, D )

A) Fail to reject the null hypothesis, there is a significant difference in performance between individuals and groups.

B) Fail to reject the null hypothesis, there is not a significant difference in performance between individuals and groups.     

C) Reject the null hypothesis, there is not a significant difference in performance between individuals and groups.

D) Reject the null hypothesis, there is a significant difference in performance between individuals and groups.


(b) Compute the estimated value of Cohen's d to measure effect size for this study. (Use 3 decimal places.)

Groups Individuals n = 12 n = 12 M = 4.0 M = 4.7 SS = 14.7 SS = 17.9

Explanation / Answer

(a) Based on these results, is there a significant difference between the performance for individuals versus the performance for groups? Use a two-tailed test with = 0.05. (Use 3 decimal places.)

t-critical =

±2.074

t =

1.408

Pooled Sd=sqrt(14.7+17.9)/22) =1.2173

Standard error = sd*sqrt(1/12+1/12) =0.497

t=(4.0-4.7)/0.497 =1.408

Df=12+12-2=22

Critical value of t at 5% level =2.074

Calculated t=1.408 < 2.074, the null hypothesis is not rejected.

Conclusion ( chose A, B, C, D )

A) Fail to reject the null hypothesis, there is a significant difference in performance between individuals and groups.

B) Fail to reject the null hypothesis, there is not a significant difference in performance between individuals and groups.     

C) Reject the null hypothesis, there is not a significant difference in performance between individuals and groups.

D) Reject the null hypothesis, there is a significant difference in performance between individuals and groups.


(b) Compute the estimated value of Cohen's d to measure effect size for this study. (Use 3 decimal places.)

Cohen's d = (4.7-4.0)/1.2173 =0.575

The effect is medium.

t-critical =

±2.074

t =

1.408