/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A study was conducted to investi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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? $$ \begin{array}{l|c|c|c|c|c|c|c|c|} \hline \text { Subject } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } \\ \hline \text { Before } & 6.6 & 6.5 & 9.0 & 10.3 & 11.3 & 8.1 & 6.3 & 11.6 \\ \hline \text { After } & 6.8 & 2.4 & 7.4 & 8.5 & 8.1 & 6.1 & 3.4 & 2.0 \\ \hline \end{array} $$

Short Answer

Expert verified
A 95% confidence interval can be constructed and used to determine if hypnotism reduces pain. If zero is not in the interval, hypnotism is effective.

Step by step solution

01

Calculate the differences

For each subject, calculate the difference between the 'before' and 'after' pain measurements. For example, for Subject A: Difference = 6.6 - 6.8 = -0.2.
02

Compute the mean difference

Sum all the differences from Step 1 and then divide by the number of subjects to find the mean difference, \(\bar{d}\).
03

Compute the standard deviation of differences

Calculate the standard deviation of the differences found in Step 1 using the formula: \[s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}\], where \({d_i}\) are the individual differences and \(n\) is the number of subjects.
04

Determine the standard error of the mean difference

Calculate the standard error of the mean difference, SE\(\bar{d}\) using: \[SE_{\bar{d}} = \frac{s_d}{\sqrt{n}}\].
05

Find the t critical value

Use a t-table to find the critical t-value (t_c) for a 95% confidence interval with \(n-1\) degrees of freedom.
06

Construct the confidence interval

Construct the \(95\%\) confidence interval for the mean difference using: \[CI = \bar{d} \pm (t_c \cdot SE_{\bar{d}})\].
07

Interpret the results

Analyze if the confidence interval includes zero. If it does not, it suggests that hypnotism may be effective in reducing pain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

hypnotism and pain reduction
Hypnotism is often explored as a method for relieving pain without the need for medication. In the context of this study, pain levels were measured before and after hypnosis to determine its effectiveness. Higher numbers on the pain scale mean more intense pain. A lower number after hypnosis implies that the technique might be working to reduce pain. By examining these changes in pain levels, researchers aim to figure out if hypnotism can statistically be proven to alleviate discomfort.
95% confidence interval
A 95% confidence interval provides a range within which we can be 95% sure that the true mean difference lies. This is based on our sample data. Here, we calculated it for the differences in pain levels before and after hypnosis. To construct this interval:
  • First, we find the mean of the differences.
  • Second, we calculate the standard deviation of those differences.
  • Next, we use these to find the standard error.
  • We then determine the appropriate t-value from the t-distribution table for our sample size.
  • Finally, we use these values to calculate the confidence interval.
This interval helps us understand if the reduction in pain after hypnosis is statistically significant.
statistical hypothesis testing
Statistical hypothesis testing allows us to make decisions or inferences about a population based on sample data. In this study:
  • The null hypothesis (H0) would be that hypnotism has no effect on pain reduction, meaning the mean difference is zero.
  • The alternative hypothesis (H1) is that hypnotism does reduce pain, meaning the mean difference is not zero.
Our goal is to use the confidence interval to see if it includes zero. If zero is not in the interval, we reject the null hypothesis. This would suggest that hypnotism does indeed have an effect on reducing pain.
paired sample t-test
A paired sample t-test is used in this study because we have two measurements from the same group of subjects – before and after hypnosis. This type of test is perfect for comparing means from the same subjects under different conditions. Steps involved:
  • Calculate the differences between each pair of observations.
  • Find the mean and standard deviation of those differences.
  • Compute the standard error using the standard deviation and the number of subjects.
  • Use the t-distribution to determine the critical t-value for our sample size.
  • Construct the confidence interval around the mean difference.
Interpretation of the paired sample t-test in this context helps determine if the observed changes from hypnotism are likely due to the treatment rather than random chance.

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Most popular questions from this chapter

Listed below are brain volumes \(\left(\mathrm{cm}^{3}\right)\) of twins from Data Set 8 "IQ and Brain Size" in Appendix B. Construct a \(99 \%\) confidence interval estimate of the mean of the differences between brain volumes for the first-born and the second-born twins. What does the confidence interval suggest? $$ \begin{array}{|l|r|r|r|r|r|r|r|r|r|r|} \hline \text { First Born } & 1005 & 1035 & 1281 & 1051 & 1034 & 1079 & 1104 & 1439 & 1029 & 1160 \\ \hline \text { Second Born } & 963 & 1027 & 1272 & 1079 & 1070 & 1173 & 1067 & 1347 & 1100 & 1204 \\ \hline \end{array} $$

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. Rhinoviruses typically cause common colds. In a test of the effectiveness of echinacea, 40 of the 45 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 88 of the 103 subjects developed rhinovirus infections (based on data from "An Evaluation of Echinacea Angustifolia in Experimental Rhinovirus Infections," by Turner et al., New England Journal of Medicine, Vol. 353, No. 4 ). We want to use a \(0.05\) significance level to test the claim that echinacea has an effect on rhinovirus infections. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, does echinacea appear to have any effect on the infection rate?

Researchers from the University of British Columbia conducted a study to investigate the effects of color on cognitive tasks. Words were displayed on a computer screen with background colors of red and blue. Results from scores on a test of word recall are given below. Higher scores correspond to greater word recall. a. Use a \(0.05\) significance level to test the claim that the samples are from populations with the same mean. b. Construct a confidence interval appropriate for the hypothesis test in part (a). What is it about the confidence interval that causes us to reach the same conclusion from part (a)? c. Does the background color appear to have an effect on word recall scores? If so, which color appears to be associated with higher word memory recall scores? $$ \begin{array}{|l|l|} \hline \text { Red Background } & n=35, \bar{x}=15.89, s=5.90 \\ \hline \text { Blue Background } & n=36, \bar{x}=12.31, s=5.48 \\ \hline \end{array} $$

20\. Does Aspirin Prevent Heart Disease? In a trial designed to test the effectiveness of aspirin in preventing heart disease, 11,037 male physicians were treated with aspirin and 11,034 male physicians were given placebos. Among the subjects in the aspirin treatment group, 139 experienced myocardial infarctions (heart attacks). Among the subjects given placebos, 239 experienced myocardial infarctions (based on data from "Final Report on the Aspirin Component of the Ongoing Physicians' Health Study," New England Journal of Medicine, Vol. 321 : \(129-135)\). Use a \(0.05\) significance level to test the claim that aspirin has no effect on myocardial infarctions. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, does aspirin appear to be effective?In a trial designed to test the effectiveness of aspirin in preventing heart disease, 11,037 male physicians were treated with aspirin and 11,034 male physicians were given placebos. Among the subjects in the aspirin treatment group, 139 experienced myocardial infarctions (heart attacks). Among the subjects given placebos, 239 experienced myocardial infarctions (based on data from "Final Report on the Aspirin Component of the Ongoing Physicians' Health Study," New England Journal of Medicine, Vol. 321 : \(129-135)\). Use a \(0.05\) significance level to test the claim that aspirin has no effect on myocardial infarctions. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, does aspirin appear to be effective?

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