Question

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

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

Given information

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

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.

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

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.

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.

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.

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.