Question

There is some evidence suggesting that you are likely to improve your test score if you rethink and change answers on a multiple-choice exam (Johnston, 1975). To examine this phenomenon, a teacher gave the same final exam to two sections of a psychology course. The students in one section were told to turn in their exams immediately after finishing, without changing any of their answers. In the other section, students were encouraged to reconsider each question and to change answers whenever they felt it was appropriate. Before the final exam, the teacher had matched 9 students in the first section with 9 students in the second section based on their midterm grades. For example, a student in the no-change section with an 89 on the midterm exam was matched with student in the change section who also had an 89 on the midterm. The difference between the two final exam grades for each matched pair was computed and the data showed that the students who were allowed to change answers scoring higher by an average of MD = 7 points with SS = 288. *PLEASE SHOW HOW TO DO WITHOUT CALCULATOR**

a. Do the data indicate a significant difference between the two conditions? Use a two-tailed test with α = .05.

b. Construct a 95% confidence interval to estimate the size of the population mean difference.

c. Write a sentence demonstrating how the results of the hypothesis test and the confidence interval would appear in a research report.

Answer 1

Answer:

(a)

Given,

To indicate significant difference between two conditions

M_D=7 , n=9, SS=288

standard deviation of difference=S_D=sqrt(SS/(n-1))

= sqrt(288/8)

= 6

By using t-test and t = M_D/(S_D/sqrt(n))

Substitute the known values in expression

=7/(6/sqrt(9))

=3.5

Degree of freedom= (n-1)

=(9-1)

df = 8

critical t(0.05,8)=2.3 is less than calculated t=3.5 so there is significant difference in two condition.

(b)

To construct a 95% confidence interval to estimate the size of population mean difference

(1-alpha)*100% confidence interval for difference = difference±t(alpha/2,n-1)*sd/sqrt(n)

= difference ± t(.05/2, n-1)*sd/sqrt(n)

Substitute the values in above formula

= 7 ± 2.3*(6/sqrt(9)

= 7±4.6

= (2.4 , 11.6)

(c)

Hence we conclude that,

there is signigicant difference in in two conditions and the we are 95% confident that the difference of scores lies between 2.4 and 11.6.