Question

Six sets of identical twins were randomly selected from a population of identical twins. One child was taken at random from each pair to form an experimental group. These children participated in a program designed to promote creative thinking. The other child from each pair was part of the control group that did not participate in the program to promote creative thinking. At the end of the program, a creative problem-solving test was given, with the results shown in the table below.

Twin Pair	A	B	C	D	E	F
Experimental group	55	33	12	25	33	47
Control Group	37	23	5	18	21	42

Higher scores indicate better performance in creative problem solving. Do the data support the claim that the program of the experimental group did promote creative problem solving? Use α = 0.01. (Let d = experimental − control.)

What are we testing in this problem?

single proportion

single mean    

difference of proportions

difference of means

paired difference

What is the level of significance?

State the null and alternate hypotheses.

What sampling distribution will you use? What assumptions are you making?

What is the value of the sample test statistic? (Round your answer to three decimal places.)

Find (or estimate) the P-value.

Sketch the sampling distribution and show the area corresponding to the P-value.

Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

(e) Interpret your conclusion in the context of the application.

There is sufficient evidence at the 0.01 to conclude that the program of the experimental group promoted creative problem solving.

There is insufficient evidence at the 0.01 to conclude that the program of the experimental group promoted creative problem solving.    

Answer 1

Ans:

paired difference

level of significance=0.01

We will use t-distribution and assume that differences of two score are normally distributed.

win Pair	A	B	C	D	E	F
Experimental group	55	33	12	25	33	47
Control Group	37	23	5	18	21	42
d	18	10	7	7	12	5
d-bar	9.833
sd	4.708

Test statistic:

t=(9.833-0)/(4.708/sqrt(6))

t=5.116

df=6-1=5

p-value=tdist(5.116,5,1)=0.0019

As,p-value<0.01,we reject the null hypothesis.

Yes,the data statistically significant at level α=0.01

There is sufficient evidence at the 0.01 to conclude that the program of the experimental group promoted creative problem solving.