Question

Male Female

16,303 24,879

25,654 13,902

1395 18,880

7767 17,119

18,717 13,158

15,903 16,953

14,586 16,267

25,740 18,634

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 Subscript dμd 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?

Identify the test statistic.

t=?

Answer 1

Sample size, n = 8.

For the score differences we have

(1) Null and Alternative Hypotheses

(2) Critical value

At = 0.05 and df = 8-1 = 7, the critical value for this left-tailed test is t_c = -1.895

(3) Test Statistics

(4) Decision about the null hypothesis

Since it is observed that t = -0.497 > t_c = -1.895, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p = 0.3171, and since p = 0.3171 > 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that males speak fewer words in a day than females, at the 0.05 significance level.