4. A study designed to measure the effect of environment on academic achievement

Question

4. A study designed to measure the effect of environment on academic achievement of 12-year old students. genetic differences may also academic achievement, the researcher wanted to control for this factor. Fifteen sets of identical twins were identified who had been adopted prior to their first birthday, with one twin placed in a home in which academics emphasized (Aca- demic) and the other placed a home in which were not emphasized (Nonacademic). The final grades (based on 100 points) for the 30 students are given here. Set of Twins 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Academic 78 75 68 92 55 74 65 80 98 52 67 55 49 66 75 Nonacademic 71 70 66 85 60 72 57 75 92 56 63 52 48 67 70 (a) Construct a 95% confidence interval for the difference between the students in academically ori- ented homes and those in nonacademically oriented homes. (b) Is there a difference in the mean final grades between the students in an academically oriented home and those in a nonacademically oriented home environment? Test at the a a 0.05 level. (Use the critical value method and not the p value method.) (c) Comment on your confidence interval in relation to your hypothesis test. Do they agree? why or why not?

Explanation / Answer

here

degree of freedom =v=((s12/n1+s22/n2)2/((s12/n1)/(n1-1)+ (s22/n2)/(n2-1))=36

also std error =(s12/n1+s22/n2)1/2 =4.102

for 36 df and 95% CI, t =2.028

hence confidence interval =(X1-X2)+/- z*std error

=(-5.3197 ; 11.3197)

as our confidence interval consist 0 as possible value we can not accept that there is a difference.

b) here t critical =+/- 2.028

and tstat =(x1-x2)/std error =0.7313

as t stat is inside the critcal region, we  can not accept that there is a difference..

c) yes they do agree, as they both depend on t critical which is fixed for both.,

S. no acad. Non acad 1 78 71 2 75 70 3 68 66 4 92 85 5 55 60 1 74 72 2 65 57 3 80 75 3 98 92 4 52 56 5 67 63 6 55 52 7 49 48 8 66 67 9 75 70 total 1049.000 1004.000 mean 69.933 66.933 std deviation(S) 14.058 11.787

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