/*! 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} Q8BSC In Exercises 5鈥16, use the lis... [FREE SOLUTION] | 魅影直播

魅影直播

In Exercises 5鈥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.

The Spoken Word Listed below are the numbers of words spoken in a day by each memberof six different couples. The data are randomly selected from the first two columns in Data Set24 鈥淲ord Counts鈥 in Appendix B.

a. Use a 0.05 significance level to test the claim that among couples, males speak fewer wordsin a day than females.

b. Construct the confidence interval that could be used for the hypothesis test described in part(a). What feature of the confidence interval leads to the same conclusion reached in part (a)?

Male

15,684

26,429

1,411

7,771

18,876

15,477

14,069

25,835

Female

24,625

13,397

18,338

17,791

12,964

16,937

16,255

18,667

Short Answer

Expert verified

a. There is not enough evidence to support the claim that males speak fewer words in a day than females.

b.The presence of 0 in the confidence interval is the feature that leads to the same conclusion as in part (a).

Step by step solution

01

Given information

The number of words spoken in a day by each member of 6 couples is recorded.

02

Hypotheses

It is claimed that males speak fewer than females.

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

Null Hypothesis: The mean of the difference between the number of words spoken by males and females is equal to 0.

\({H_0}:{\mu _d} = 0\)

Alternative Hypothesis: The mean difference between the number of words spoken by males and females is less than 0.

\({H_1}:{\mu _d} < 0\)

The test is left-tailed.

03

Differences in the values of each matched pair

The following table shows the differences in the number of words spoken by males and females for each matched pair:

Male

15684

26429

1411

7771

18876

15477

14069

25835

Female

24625

13397

18338

17791

12964

16937

16255

18667

Differences(d)

-8941

13032

-16927

-10020

5912

-1460

-2186

7168

04

Mean and standard deviation of the differences

The number of pairs (n) is equal to 8.

The mean value of the differences is computed below:

\(\begin{array}{c}\bar d = \frac{{\left( { - 8941} \right) + 13032 + ...... + 7168}}{8}\\ = - 1678\end{array}\)

The standard deviation of the differences is computed below:

\(\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({d_i} - \bar d)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( { - 8941 - \left( { - 1678} \right)} \right)}^2} + {{\left( {13032 - \left( { - 1678} \right)} \right)}^2} + ....... + {{\left( {7168 - \left( { - 1678} \right)} \right)}^2}}}{{8 - 1}}} \\ = 10052.87\end{array}\)

The mean value of the differences for the population of matched pairs \(\left( {{\mu _d}} \right)\) is considered to be equal to 0.

05

Test Statistic

The value of the test statistic is computed as shown:

\(\begin{array}{c}t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{ - 1678 - 0}}{{\frac{{10052.87}}{{\sqrt 8 }}}}\\ = - 0.472\end{array}\)

The degrees of freedom are computed below:

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

Referring to the t-distribution table, the critical value of t at\(\alpha = 0.05\)and degrees of freedom equal to 7 for a left-tailed test is equal to -1.8946.

The left-tailed p-value for t equal to -0.472 is equal to 0.3256.

06

Conclusion of the test

a.

Since the test statistic value is greater than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to support the claim that males speak fewer words in a day than females.

07

Construct the confidence interval

b.

The confidence interval has the following expression:

\(CI = \bar d - E < {\mu _d} < \bar d + E\)

The value of the margin of error (E) is given by the formula:

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

The confidence level to construct the confidence interval for a one-tailed test\(\alpha = 0.05\)is equal to 90%.

Thus, the value of\(\alpha \)for constructing the confidence interval is equal to 0.

The value of\({t_{\frac{\alpha }{2}}}\)for 7 degrees of freedom when\(\alpha = 0.10\)is equal to 1.8946.

Substitute the respective values to compute the margin of error.

\(\begin{array}{c}E = 1.8946 \times \frac{{10052.87}}{{\sqrt 8 }}\\ = 6733.84\end{array}\)

Substitute the required values in the formula to compute the confidence interval.

\(\begin{array}{c} - 1678 - 6733.84 < {\mu _d} < - 1678 + 6733.84\\ - 8412 < {\mu _d} < 5056\end{array}\)

It can be observed that 0 lies within the interval. This implies that the difference in the number of words spoken in a day by men and women can be equal to 0.

Thus it can be concluded that there is not enough evidence to support the claim that males speak fewer words in a day than females.

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 魅影直播!

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

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 鈥淓yewitness Memory of Police Trainees for Realistic Role Plays,鈥 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 鈥渟tress decreases the amount recalled.鈥

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

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

In Exercises 5鈥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 鈥淭able鈥 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 鈥淐ourse Evaluations鈥 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?

Testing Claims About Proportions. In Exercises 7鈥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

Question:Headache Treatment In a study of treatments for very painful 鈥渃luster鈥 headaches, 150 patients were treated with oxygen and 148 other patients were given a placebo consisting of ordinary air. Among the 150 patients in the oxygen treatment group, 116 were free from head- aches 15 minutes after treatment. Among the 148 patients given the placebo, 29 were free from headaches 15 minutes after treatment (based on data from 鈥淗igh-Flow Oxygen for Treatment of Cluster Headache,鈥 by Cohen, Burns, and Goads by, Journal of the American Medical Association, Vol. 302, No. 22). We want to use a 0.01 significance level to test the claim that the oxygen treatment is effective.

a. Test the claim using a hypothesis test.

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

c. Based on the results, is the oxygen treatment effective?

In Exercises 5鈥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 鈥淭able鈥 answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).)

IQ and Lead Exposure Data Set 7 鈥淚Q and Lead鈥 in Appendix B lists full IQ scores for a random sample of subjects with low lead levels in their blood and another random sample of subjects with high lead levels in their blood. The statistics are summarized below.

a. Use a 0.05 significance level to test the claim that the mean IQ score of people with low blood lead levels is higher than the mean IQ score of people with high blood lead levels.

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

c. Does exposure to lead appear to have an effect on IQ scores?

Low Blood Lead Level: n = 78, \(\bar x\) = 92.88462, s = 15.34451

High Blood Lead Level: n = 21,\(\bar x\)= 86.90476, s = 8.988352

True?For the methods of this section, which of the following statements are true?

a.When testing a claim with ten matched pairs of heights, hypothesis tests using the P-valuemethod, critical value method, and confidence interval method will all result in the same conclusion.

b.The methods of this section are robustagainst departures from normality, which means that the distribution of sample differences must be very close to a normal distribution.

c.If we want to use a confidence interval to test the claim that\({{\bf{\mu }}_{\bf{d}}}{\bf{ < 0}}\)with a 0.01 significancelevel, the confidence interval should have a confidence level of 98%.

d.The methods of this section can be used with annual incomes of 50 randomly selected attorneysin North Carolina and 50 randomly selected attorneys in South Carolina.

e.With ten matched pairs of heights, the methods of this section require that we use n= 20.

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.