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 35 14 25 33 47 Control Group 39 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 ? = 0.01. (Let d = experimental − control.) What are we testing in this problem? paired difference or difference of means Or single mean or difference of proportions or single proportion What is the level of significance? State the null and alternate hypotheses. H0: ?d = 0; H1: ?d ≠ 0 H0: ?d > 0; H1: ?d ≤ 0 H0: ?d ≠ 0; H1: ?d = 0 H0: ?d ≤ 0; H1: ?d > 0 What sampling distribution will you use? What assumptions are you making? a.The standard normal. We assume that d has an approximately normal distribution. b.The Student's t. We assume that d has an approximately normal distribution. C. The Student's t. We assume that d has an approximately uniform distribution.The standard normal. D.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 > 0.250 0.125 < P-value < 0.250 0.050 < P-value < 0.125 0.025 < P-value < 0.050 0.005 < P-value < 0.025 P-value < 0.005 Will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?? A.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant b.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. C.At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant. D.At the ? = 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. A.There is sufficient evidence at the 0.01 to conclude that the program of the experimental group promoted creative problem solving. b.There is insufficient evidence at the 0.01 to conclude that the program of the experimental group promoted creative problem solving.

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 35 14 25 33 47 Control Group 39 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 ? = 0.01. (Let d = experimental − control.) What are we testing in this problem? paired difference or difference of means Or single mean or difference of proportions or single proportion What is the level of significance? State the null and alternate hypotheses. H0: ?d = 0; H1: ?d ≠ 0 H0: ?d > 0; H1: ?d ≤ 0 H0: ?d ≠ 0; H1: ?d = 0 H0: ?d ≤ 0; H1: ?d > 0 What sampling distribution will you use? What assumptions are you making? a.The standard normal. We assume that d has an approximately normal distribution. b.The Student's t. We assume that d has an approximately normal distribution. C. The Student's t. We assume that d has an approximately uniform distribution.The standard normal. D.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 > 0.250 0.125 < P-value < 0.250 0.050 < P-value < 0.125 0.025 < P-value < 0.050 0.005 < P-value < 0.025 P-value < 0.005 Will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?? A.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant b.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. C.At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant. D.At the ? = 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. A.There is sufficient evidence at the 0.01 to conclude that the program of the experimental group promoted creative problem solving. b.There is insufficient evidence at the 0.01 to conclude that the program of the experimental group promoted creative problem solving.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Twin Pair	A	B	C	D	E	F
Experimental group	55	35	14	25	33	47
Control Group	39	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 ? = 0.01. (Let d = experimental − control.)

What are we testing in this problem?

paired difference or difference of means Or single mean or difference of proportions or single proportion

What is the level of significance?

State the null and alternate hypotheses.

H₀: ?_d = 0; H₁: ?_d ≠ 0

H₀: ?_d > 0; H₁: ?_d ≤ 0

H₀: ?_d ≠ 0; H₁: ?_d = 0

H₀: ?_d ≤ 0; H₁: ?_d > 0

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

a.The standard normal. We assume that d has an approximately normal distribution.

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

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

D.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 > 0.250

0.125 < P-value < 0.250

0.050 < P-value < 0.125

0.025 < P-value < 0.050

0.005 < P-value < 0.025

P-value < 0.005

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

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

b.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.

C.At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

D.At the ? = 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.

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

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

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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