In Exercises 5–16, use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal.

Hypnotism for Reducing Pain A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are given in the accompanying table (based on “An Analysis of Factors That Contribute to the Efficacy of Hypnotic Analgesia,” by Price and Barber, Journal of Abnormal Psychology, Vol. 96, No. 1). The values are before and after hypnosis; the measurements are in centimeters on a pain scale. Higher values correspond to greater levels of pain. Construct a 95% confidence interval for the mean of the “before/after” differences. Does hypnotism appear to be effective in reducing pain?

Subject

A

B

C

D

E

F

G

H

Before

6.6

6.5

9

10.3

11.3

8.1

6.3

11.6

After

6.8

2.4

7.4

8.5

8.1

6.1

3.4

2

Short Answer

Expert verified

The 95% confidence interval for the mean of the differences between the pain measurements before hypnosis and after hypnosis is equal to (0.69 cm,5.56 cm).

It seems that there are only positive values in the interval, and the value of 0 does not lie in the interval.

Therefore, theeffect of hypnotism is significant in reducing pain.

Step by step solution

01

Given information

The given data is based on the use of hypnotism for pain reduction. The measurements are in centimetreson a pain scale. The measurements are obtained before hypnosis and after hypnosis.

02

Hypothesis for problem

Null Hypothesis: The mean of the differences between the pain measurements before and after hypnosis is equal to 0.

\({H_{0\;}}:\;{\mu _d} = 0\)

Alternative Hypothesis:The mean of the differences between the pain measurements before and after hypnosis is greater than 0.\({H_1}\;:{\mu _d} > 0\;\..

03

Expression of the confidence interval

The formula of the confidence interval is as follows:

\({\rm{C}}{\rm{.I}} = \bar d - E < {\mu _d} < \bar d + E\;\)

The value of the margin of error (E) has the following notation:

\(E = {t_{\frac{\alpha }{2},df}} \times \frac{{{s_d}}}{{\sqrt n }}\)

04

Table of the differences

The following table shows the difference in the pain measurements before and after hypnosis:

Subject

A

B

C

D

E

F

G

H

Before

6.6

6.5

9

10.3

11.3

8.1

6.3

11.6

After

6.8

2.4

7.4

8.5

8.1

6.1

3.4

2

Differences (d)

-0.2

4.1

1.6

1.8

3.2

2

2.9

9.6

05

Find the mean of the differences

Now, the mean of the differences between the two pain measurements is computed below:

\(\begin{array}{c}\bar d = \frac{{\sum\limits_{i = 1}^n {{d_i}} }}{n}\\ = \frac{{\left( { - 0.2} \right) + 4.1 + \cdots + 9.6}}{8}\\ = 3.125\end{array}\)

06

Find the standard deviation of the differences

The standard deviation of the differences is equal to:

\(\begin{array}{c}s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{d_i} - \bar d} \right)}^2}} }}{{n - 1}}} \;\\ = \sqrt {\frac{{{{\left( { - 0.2 - 3.125} \right)}^2} + {{\left( {4.1 - 3.125} \right)}^2} + \cdots + {{\left( {9.6 - 3.125} \right)}^2}}}{{8 - 1}}} \\ = 2.911\end{array}\)

07

Find the critical value

The value of the degrees of freedom is equal to:

\(\begin{array}{c}{\rm{df}} = n - 1\\ = 8 - 1\\ = 7\end{array}\)

The confidence level is equal to 95%. Thus, the level of significance becomes 0.05.

Therefore,

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025\end{array}\)

Referring to the t-distribution table, the critical value of t for 7 degrees of freedom at 0.025 significance level is equal to 2.3646.

08

Compute the margin of error

The value of the margin of error is computed below:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2},df}} \times \frac{{{s_d}}}{{\sqrt n }}\\ = {t_{\frac{{0.05}}{2},7}} \times \frac{{2.911}}{{\sqrt 8 }}\\ = 2.3646 \times \frac{{2.911}}{{\sqrt 8 }}\\ = 2.433992\end{array}\)

09

Step 9:Computethe confidence interval

The value of the confidence interval is equal to:

\(\begin{array}{c}\bar d - E < {\mu _d} < \bar d + E\;\\\left( {3.125 - 2.433992} \right) < {\mu _d} < \left( {3.125 + 2.433992} \right)\\0.69 < {\mu _d} < 5.56\end{array}\)

Thus, the 95% confidence interval is equal to (0.69 cm,5.56 cm).

10

Interpretation of the confidence interval

It can be seen that the value of 0 is not included in the interval, and all the values are positive.

This implies that the mean of the differences between the pain measurements cannot be equal to 0.

And, the pain measurement after hypnosis is less than the pain measurement before hypnosis.

Thus, there is enough evidence to conclude that hypnotism appears to be effective in reducing pain.

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Most popular questions from this chapter

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\))

Regular Coke and Diet Coke Data Set 26 “Cola Weights and Volumes” in Appendix B includesweights (lb) of the contents of cans of Diet Coke (n= 36,\(\overline x \)= 0.78479 lb, s= 0.00439 lb) and of the contents of cans of regular Coke (n= 36,\(\overline x \)= 0.81682 lb, s= 0.00751 lb).

a. Use a 0.05 significance level to test the claim that the contents of cans of Diet Coke have weights with a mean that is less than the mean for regular Coke.

b. Construct the confidence interval appropriate for the hypothesis test in part (a).

c. Can you explain why cans of Diet Coke would weigh less than cans of regular Coke?

Testing Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Dreaming in Black and White A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 306 people over the age of 55, 68 dream in black and white, and among 298 people under the age of 25, 13 dream in black and white (based on data from “Do We Dream in Color?” by Eva Murzyn, Consciousness and Cognition, Vol. 17, No. 4). We want to use a 0.01 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion of those under 25.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. An explanation given for the results is that those over the age of 55 grew up exposed to media that was mostly displayed in black and white. Can the results from parts (a) and (b) be used to verify that explanation?

Heights Use a 0.01 significance level with the sample data from Exercise 3 to test the claim that women have heights with a mean that is less than the mean height of men.

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).)

BMI We know that the mean weight of men is greater than the mean weight of women, and the mean height of men is greater than the mean height of women. A person’s body mass index (BMI) is computed by dividing weight (kg) by the square of height (m). Given below are the BMI statistics for random samples of females and males taken from Data Set 1 “Body Data” in Appendix B.

a. Use a 0.05 significance level to test the claim that females and males have the same mean BMI.

b. Construct the confidence interval that is appropriate for testing the claim in part (a).

c. Do females and males appear to have the same mean BMI?

Female BMI: n = 70, \(\bar x\) = 29.10, s = 7.39

Male BMI: n = 80, \(\bar x\) = 28.38, s = 5.37

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of \({n_1} - 1\) and \({n_2} - 1\).)

Second-Hand Smoke Data Set 12 “Passive and Active Smoke” in Appendix B includes cotinine levels measured in a group of nonsmokers exposed to tobacco smoke (n = 40, \(\bar x\) = 60.58 ng/mL, s = 138.08 ng/mL) and a group of nonsmokers not exposed to tobacco smoke (n = 40, \(\bar x\) = 16.35 ng/mL, s = 62.53 ng/mL). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced.

a. Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke.

b. Construct the confidence interval appropriate for the hypothesis test in part (a).

c. What do you conclude about the effects of second-hand smoke?

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