Question

listed below are the numbers of words spoken in a day by each member of eight different randomly selected couples. complete parts (a) and (b) below
male: 15,178; 27,485; 1409; 7958; 18,497; 15,294; 13,960; 26,335
female: 25,243; 13,139; 17,986; 17,387; 12,678; 16,310; 16,744; 18,183

a. use a 0.05 significance level to test the claim that among couples, males speak fewer words in a day than females.
in this example, ( mu_d ) is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the words spoken by the male minus words spoken by the female. what are the null and alternative hypotheses for the hypothesis test?
( h_0: mu_d = 0 ) (type integers or decimals. do not round.)
( h_1: mu_d < 0 )
identify the test statistic.
( t = -0.39 ) (round to two decimal places as needed.)
identify the p - value.
p - value = 0.354 (round to three decimal places as needed.)
what is the conclusion based on the hypothesis test?
since the p - value is greater than the significance level, fail to reject the null hypothesis. there is not sufficient evidence to support the claim that males speak fewer words in 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)?
the confidence interval is ( -9407 < mu_d < 6774 ) (round to the nearest integer as needed.)

Explanation:

Step1: Define Hypotheses

The null hypothesis \( H_0 \) is that the mean difference \( \mu_d = 0 \) (no difference in words spoken by males and females). The alternative hypothesis \( H_1 \) is \( \mu_d < 0 \) (males speak fewer words, so male - female difference is negative).

Step2: Calculate Differences

First, find the differences \( d = \text{Male words}-\text{Female words} \) for each couple:

• \( 15178 - 25243=-10065 \)
• \( 27485 - 13139 = 14346 \)
• \( 1409 - 17986=-16577 \)
• \( 7958 - 17387=-9429 \)
• \( 18497 - 12678 = 5819 \)
• \( 15294 - 16310=-1016 \)
• \( 13960 - 16744=-2784 \)
• \( 26335 - 18183 = 8152 \)

Step3: Compute Mean and Std Dev of Differences

Calculate the mean of differences \( \bar{d} \) and standard deviation \( s_d \). Using a calculator or software, we find \( \bar{d}\approx - 1413.25 \) and \( s_d\approx 9244.42 \) (sample size \( n = 8 \)).

Step4: Calculate Test Statistic

The test statistic for a paired - t - test is \( t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}} \). Here, \( \mu_d = 0 \), so \( t=\frac{-1413.25}{9244.42/\sqrt{8}}\approx - 0.39 \) (rounded to two decimal places).

Step5: Find P - value

Using a t - distribution with \( df=n - 1=7 \) and the test statistic \( t=-0.39 \), the P - value for a left - tailed test is the probability that \( T < - 0.39 \) where \( T\sim t(7) \). Using a t - table or software, the P - value is approximately \( 0.354 \) (rounded to three decimal places).

Step6: Make Conclusion

Compare the P - value (\( 0.354 \)) with the significance level \( \alpha = 0.05 \). Since \( 0.354>0.05 \), we fail to reject \( H_0 \). There is not sufficient evidence to support the claim that males speak fewer words.

Step7: Confidence Interval for \( \mu_d \)

For a left - tailed test with \( \alpha = 0.05 \), the confidence level for the interval is \( 90\% \) (since for a one - tailed test, the confidence interval corresponding to the hypothesis test has confidence level \( 1 - 2\alpha \) for two - tailed, but for one - tailed, we use \( 1-\alpha \) for the relevant side). The formula for the confidence interval for \( \mu_d \) is \( \bar{d}\pm t_{\alpha/2,df}\frac{s_d}{\sqrt{n}} \). For \( df = 7 \) and \( \alpha=0.05 \) (one - tailed, so \( t_{\alpha,df}=t_{0.05,7}\approx 1.895 \)). The lower and upper bounds of the confidence interval for \( \mu_d < 0 \) - related interval (since our alternative is \( \mu_d < 0 \)) can be calculated. The interval is \( - 9407<\mu_d<6774 \) (rounded to nearest integer), and since \( 0 \) is within this interval, we fail to reject \( H_0 \), which matches the hypothesis test conclusion.

Answer:

a. \( H_0:\mu_d = 0 \), \( H_1:\mu_d < 0 \); Test statistic \( t=-0.39 \); P - value \( = 0.354 \); Since P - value \(>0.05 \), fail to reject \( H_0 \), not sufficient evidence.
b. Confidence interval \( - 9407<\mu_d<6774 \); \( 0 \) is in the interval, so same conclusion as part (a).