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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.

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 “Word Counts” 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.

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Most popular questions from this chapter

A sample size that will ensure a margin of error of at most the one specified.

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\({n_2} - 1\).)

IQ and Lead Exposure Data Set 7 “IQ and Lead” 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

Repeat Exercise 12 “IQ and Lead” by assuming that the two population standard deviations are equal, so \({\sigma _1} = {\sigma _2}\). Use the appropriate method from Part 2 of this section. Does pooling the standard deviations yield results showing greater significance?

Hypothesis and conclusions refer to the hypothesis test described in exercise 1.

a. Identify the null hypothesis and alternative hypothesis

b. If the p-value for test is reported as “less than 0.001,” what should we conclude about the original claim?

In Exercises 5–20, assume that the two samples are independent random samples selected from normally distributed populations. 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 n1−1 and n2−1).

Are male and female professors rated differently? According to Data Set 17 “Course Evaluations” Appendix B, given below are student evaluation scores of female professors and male professors. The test claims that female and male professors have the same mean evaluation ratings. Does there appear to be a difference?

Females

4.4

3.4

4.8

2.9

4.4

4.9

3.5

3.7

3.4

4.8

Males

4.0

3.6

4.1

4.1

3.5

4.6

4.0

4.3

4.5

4.3

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