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 comma 375 26 comma 556 1401 7559 18 comma 327 15 comma 076 13 comma 760 26 comma 649 Female 24 comma 523 13 comma 931 18 comma 999 17 comma 770 13 comma 431 17 comma 388 16 comma 728 18 comma 102 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 Subscript 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? Upper H 0​: mu Subscript d equals 0 ​word(s) Upper H 1​: mu Subscript d less than 0 ​word(s) ​(Type integers or decimals. Do not​ round.) Identify the test statistic. tequals nothing ​(Round to two decimal places as​ needed.)

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d = 0

Alternative hypothesis: u_d < 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 10250.627

SE = s / sqrt(n)

S.E = 3624.144

DF = n - 1 = 8 -1

D.F = 7

t = [ (x₁ - x₂) - D ] / SE

t = - 0.56

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 7 degrees of freedom is less than - 0.56.

Thus, the P-value = 0.296

Interpret results. Since the P-value (0.296) is greater than the significance level (0.05), we have to accept the null hypothesis.

Do not reject H0, we have sufficient evidence in the favor of the claim that among​ couples, males speak fewer words in a day than females.