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	53	33	12	25	33	47
Control Group	37	23	7	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 proportionpaired difference    difference of proportionsdifference of meanssingle mean

What is the level of significance?


State the null and alternate hypotheses.

H₀: μ_d = 0; H₁: μ_d ≠ 0H₀: μ_d > 0; H₁: μ_d ≤ 0    H₀: μ_d ≤ 0; H₁: μ_d > 0H₀: μ_d ≠ 0; H₁: μ_d = 0

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

The Student's t. We assume that d has an approximately uniform distribution.The standard normal. We assume that d has an approximately normal distribution.    The Student's t. We assume that d has an approximately normal distribution.The standard normal. We assume that d has an approximately uniform distribution.

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


Estimate the P-value.

P-value > 0.2500.125 < P-value < 0.250    0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005

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

Will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.    At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

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

paired difference  

level of significance =0.01

H₀: μ_d ≤ 0; H₁: μ_d > 0

The Student's t. We assume that d has an approximately normal distribution.

S. No	experimental	Control	diff:(d)=x1-x2	d2
1	53	37	16	256.00
2	33	23	10	100.00
3	12	7	5	25.00
4	25	18	7	49.00
5	33	21	12	144.00
6	47	42	5	25.00
	total	=	Σd=55	Σd2=599
	mean dbar=	d̅     =	9.167
	degree of freedom =n-1                            =		5.000
	Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) =		4.36
	std error=Se=SD/√n=		1.7780
	test statistic            =	(d̅-μd)/Se         =	5.156

P-value < 0.005

At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.

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