Question

We are interested in conducting a study to determine whether a study intervention significantly improves college students’ test scores. The population’s GPA population mean= 2.64 and standard deviation= 0.42. We hope that intervention will improve girls’ body esteem scores by 0.25 points. We plan to recruit a sample of 36 students. We set our alpha level at .05 for two-tailed test. Determine the statistical power that this proposed study would have; report your answer as a percentage. Show ALL of your calculations for full credit.

Answer 1

Answer:

GIven,

Ho : u = 2.64

Ha : u != 2.64

Here at 95% CI, z value is 1.96

consider,

Interval = xbar +/- z*s/sqrt(n)

substitute values

= 2.64 +/- 1.96*0.42/sqrt(36)

= (2.7772 , 2.5028)

Here the score is increased by 0.25 times

then,

mean = u = 2.64 + 0.25 = 2.89

Now mean = 2.89

standard deviation = s/sqrt(n) = 0.42/sqrt(36) = 0.07

Power = P(X <= 2.5028) + P(X >= 2.7772)

= P(z < (2.5028 - 2.89)/0.07) + P(z > (2.7772 - 2.89)/0.07)

= P(z < -5.5314) + P(z > -1.6114)

= 0 + 0.9464537 [since from z table]

= 0.9465

= 94.65%