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: 16,100 25,783 1424 7915 18,978 15,485 13,977 26,757
female: 25,435 13,880 18,267 18,193 12,868 17,080 16,236 19,160
3 homework
question 1, 9.3.6-t
part 6 of 6
hw score: 32.88%, 2.63 of 8 points
points: 0.73 of 1
ny statistics shariati f25

a. use a 0.01 significance level to test the claim that among couples, males speak fewer words in a day than females.
in this example, ( h_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: h_d = 0 ) (words)
( h_1: h_d < 0 ) (words)
(type integers or decimals. do not round.)
identify the test statistic.
( t = -0.52 ) (round to two decimal places as needed.)
identify the p - value.
p - value = 0.309 (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 (round to the nearest integer as needed.)( h_d < ) (words)

Explanation:

Step1: Identify Hypotheses

The null hypothesis \( H_0 \) assumes no difference, so \( H_0: \mu_d = 0 \). The alternative \( H_1 \) tests if males speak fewer words, so \( H_1: \mu_d < 0 \) (where \( d = \text{male words} - \text{female words} \)).

Step2: Test Statistic

Given \( t = -0.52 \) (from the problem, likely calculated via paired t - test formula \( t=\frac{\bar{d}- \mu_d}{s_d/\sqrt{n}} \), but here we use the provided value).

Step3: P - value Interpretation

P - value = 0.309. Since \( 0.309>0.01 \) (significance level), we fail to reject \( H_0 \). There's not enough evidence males speak fewer words.

Step4: Confidence Interval Feature

For a left - tailed test with \( \alpha = 0.01 \), a 99% confidence interval (or 98%? Wait, for one - tailed \( \alpha = 0.01 \), two - tailed confidence level is \( 1 - 2\alpha= 0.98 \)? Wait, no: for a one - tailed test at \( \alpha = 0.01 \), the confidence interval for \( \mu_d \) should be constructed such that if 0 is not in the interval (for upper or lower tail), but here since we failed to reject, 0 is in the confidence interval. The key feature is that the confidence interval contains 0, leading to the same conclusion (fail to reject \( H_0 \)) as the hypothesis test.

Answer:

a. \( H_0: \mu_d = 0 \), \( H_1: \mu_d < 0 \); Test statistic \( t=-0.52 \); P - value = 0.309; Conclusion: Fail to reject \( H_0 \) (not enough evidence males speak fewer words).
b. The confidence interval contains 0, leading to the same conclusion as in (a).