Eyewitness Accuracy of Police Does stress affect the recall ability of police eyewitnesses? This issue was studied in an experiment that tested eyewitness memory a week after a nonstressful interrogation of a cooperative suspect and a stressful interrogation of an uncooperative and belligerent suspect. The numbers of details recalled a week after the incident were recorded, and the summary statistics are given below (based on data from “Eyewitness Memory of Police Trainees for Realistic Role Plays,” by Yuille et al., Journal of Applied Psychology, Vol. 79, No. 6). Use a 0.01 significance level to test the claim in the article that “stress decreases the amount recalled.”

Nonstress: n = 40,\(\bar x\)= 53.3, s = 11.6

Stress: n = 40,\(\bar x\)= 45.3, s = 13.2

Short Answer

Expert verified

There is enough evidence to conclude that stress decreases the amount of information recalled by an eyewitness.

Step by step solution

01

Given information

The mean value, sample size and standard deviation values are given for the amount of information recalled by 2 samples of eyewitnesses after a non-stressful interrogation and after a stressful interrogation.

02

Hypotheses

It is claimed that stress decreases the amount of information recalled by an eyewitness.

Corresponding to the given claim, the following hypotheses are set up:

Null Hypothesis: The amount of information recalled after a non-stressful interrogation is equal to the amount of information recalled after a stressful interrogation.

\({H_0}:{\mu _1} = {\mu _2}\)

Alternative Hypothesis: The amount of information recalled after a stressful interrogation is less than the amount of information recalled after a non-stressful interrogation.

\({H_1}:{\mu _1} > {\mu _2}\)

The test is right-tailed.

03

Sample sizes, sample means and pooled variance

Sample Means:

Let\({\bar x_1}\)denote the mean amount of information recalled after a non-stressful interrogation.

The value of\({\bar x_1}\)is equal to 53.3.

Let\({\bar x_2}\)denote the mean amount of information recalled after a stressful interrogation.

The value of\({\bar x_2}\)is equal to 45.3.

Sample Sizes:

Let\({n_1}\)denote the sample size corresponding to the non-stressful interrogation.

The value of\({n_1}\)is equal to 40.

Let\({n_2}\)denote the sample size corresponding to the stressful interrogation.

The value of\({n_2}\)is equal to 40.

Pooled Variance:

Let\({s_1}\)denote the standard deviation of the amount of information recalled after a non-stressful interrogation.

The value of\({s_1}\)is equal to 11.6.

Let\({s_2}\)denote the standard deviation of the amount of information recalled after a stressful interrogation.

The value of\({s_2}\)is equal to 13.2.

The value of the pooled variance is computed below:

\(\begin{aligned} s_p^2 &= \frac{{\left( {{n_1} - 1} \right)s_1^2 + \left( {{n_2} - 1} \right)s_2^2}}{{\left( {{n_1} - 1} \right) + \left( {{n_2} - 1} \right)}}\\ &= \frac{{\left( {40 - 1} \right){{\left( {11.6} \right)}^2} + \left( {40 - 1} \right){{\left( {13.2} \right)}^2}}}{{\left( {40 - 1} \right) + \left( {40 - 1} \right)}}\\ &= 154.4\end{aligned}\)

04

Calculate test statistic, critical value and p-value

The test statistic value is computed as follows:

\(\begin{aligned} t &= \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_p^2}}{{{n_1}}} + \frac{{s_p^2}}{{{n_2}}}} }}\;\;\;{\rm{where}}\;{\mu _1} - {\mu _2}\;is\;0\\ &= \frac{{\left( {53.3 - 45.3} \right) - 0}}{{\sqrt {\frac{{154.4}}{{40}} + \frac{{154.4}}{{40}}} }}\\ &= 2.879\end{aligned}\)

The value of the degrees of freedom is equal to:

\(\begin{aligned} df &= {n_1} + {n_2} - 2\\ &= 40 + 40 - 2\\ &= 78\end{aligned}\)

Referring to t-table:

The critical value of t for \(\alpha = 0.01\) and 78 degrees of freedom for a right-tailed test is equal to 2.3751.

The corresponding p-value is equal to 0.0026.

Since the test statistic (2.879) is greater than the critical value and the p-value is less than 0.01, the null hypothesis is rejected.

05

Conclusion

There is enough evidence to conclude thatstress decreases the amount of information recalled by an eyewitness.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

Smoking Cessation Programs Among 198 smokers who underwent a “sustained care” program, 51 were no longer smoking after six months. Among 199 smokers who underwent a “standard care” program, 30 were no longer smoking after six months (based on data from “Sustained Care Intervention and Postdischarge Smoking Cessation Among Hospitalized Adults,” by Rigotti et al., Journal of the American Medical Association, Vol. 312, No. 7). We want to use a 0.01 significance level to test the claim that the rate of success for smoking cessation is greater with the sustained care program.

a. Test the claim using a hypothesis test.

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

c. Does the difference between the two programs have practical significance?

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.

Cell Phones and Handedness A study was conducted to investigate the association between cell phone use and hemispheric brain dominance. Among 216 subjects who prefer to use their left ear for cell phones, 166 were right-handed. Among 452 subjects who prefer to use their right ear for cell phones, 436 were right-handed (based on data from “Hemi- spheric Dominance and Cell Phone Use,” by Seidman et al., JAMA Otolaryngology—Head & Neck Surgery, Vol. 139, No. 5). We want to use a 0.01 significance level to test the claim that the rate of right-handedness for those who prefer to use their left ear for cell phones is less than the rate of right-handedness for those who prefer to use their right ear for cell phones. (Try not to get too confused here.)

a. Test the claim using a hypothesis test.

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

Determining Sample Size The sample size needed to estimate the difference between two population proportions to within a margin of error E with a confidence level of 1 - a can be found by using the following expression:

\({\bf{E = }}{{\bf{z}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}}\sqrt {\frac{{{{\bf{p}}_{\bf{1}}}{{\bf{q}}_{\bf{1}}}}}{{{{\bf{n}}_{\bf{1}}}}}{\bf{ + }}\frac{{{{\bf{p}}_{\bf{2}}}{{\bf{q}}_{\bf{2}}}}}{{{{\bf{n}}_{\bf{2}}}}}} \)

Replace \({{\bf{n}}_{\bf{1}}}\;{\bf{and}}\;{{\bf{n}}_{\bf{2}}}\) by n in the preceding formula (assuming that both samples have the same size) and replace each of \({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{q}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}\;{\bf{and}}\;{{\bf{q}}_{\bf{2}}}\)by 0.5 (because their values are not known). Solving for n results in this expression:

\({\bf{n = }}\frac{{{\bf{z}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}^{\bf{2}}}}{{{\bf{2}}{{\bf{E}}^{\bf{2}}}}}\)

Use this expression to find the size of each sample if you want to estimate the difference between the proportions of men and women who own smartphones. Assume that you want 95% confidence that your error is no more than 0.03.

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\({n_2} - 1\).)

Are Male Professors and Female Professors Rated Differently?

a. Use a 0.05 significance level to test the claim that two samples of course evaluation scores are from populations with the same mean. Use these summary statistics: Female professors:

n = 40, \(\bar x\)= 3.79, s = 0.51; male professors: n = 53, \(\bar x\) = 4.01, s = 0.53. (Using the raw data in Data Set 17 “Course Evaluations” will yield different results.)

b. Using the summary statistics given in part (a), construct a 95% confidence interval estimate of the difference between the mean course evaluations score for female professors and male professors.

c. Example 1 used similar sample data with samples of size 12 and 15, and Example 1 led to the conclusion that there is not sufficient evidence to warrant rejection of the null hypothesis.

Do the larger samples in this exercise affect the results much?

Degrees of Freedom

For Example 1 on page 431, we used df\( = \)smaller of\({n_1} - 1\)and\({n_2} - 1\), we got\(df = 11\), and the corresponding critical values are\(t = \pm 2.201.\)If we calculate df using Formula 9-1, we get\(df = 19.063\), and the corresponding critical values are\( \pm 2.093\). How is using the critical values of\(t = \pm 2.201\)more “conservative” than using the critical values of\( \pm 2.093\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free