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	16,054	27,338	1411	8019	19,507	15,326	13,597	25,343
Female	25,186	13,173	18,067	17,119	13,192	17,360	16,313	18,554

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,

μd

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?

H0​:

μd

equals=

00 ​word(s)

H1​:

μd

less than<

00 ​word(s)

​(Type integers or decimals. Do not​ round.)

Identify the test statistic.

t=nothing

​(Round to two decimal places as​ needed.)

Identify the​ P-value.

​P-value=nothing

​(Round to three decimal places as​ needed.)

What is the conclusion based on the hypothesis​ test?

Since the​ P-value is

▼

greater than

less than or equal to

the significance​ level,

▼

fail to reject

reject

the null hypothesis. There

▼

is not

is

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

nothing

​word(s)<μd<nothing

​word(s).

​(Round to the nearest integer as​ needed.)

What feature of the confidence interval leads to the same conclusion reached in part​ (a)?

Since the confidence interval contains

▼

zero,

only negative numbers,

only positive numbers,

▼

reject

fail to reject

the null hypothesis.

Answer 1

Assuming that the data given in the question is as follows :

Male	Female	Difference (D)
16,054	25,186	-9,132
27,338	13,173	14,165
1411	18,067	-16,656
8019	17,119	-9,100
19,507	13,192	6,315
15,326	17,360	-2,034
13,597	16,313	-2,716
25,343	18,554	6,789

Thus,  $\bar{x}_{D}$ = -1546.125, S_D = 10160.37487, n = 8

Ho :  $\mu_{D}$ = 0
Ho :  $\mu_{D}$ < 0

Level of Significance(l.o.s.) :  $\alpha$ = 0.05

Decision Criteria : Reject Ho at 5% l.o.s. if t cal < t tab,
where t tab = t ($\alpha$, n-1) = t (0.05,7) = - 1.8946

Calculation :

t cal =  $\frac{\bar{x}_{D}-\mu_{D}}{S_{D}/\sqrt{n}}$ =  $\frac{-1546.125-0}{10160.37487/\sqrt{8}}$ = -0.43041 = -0.43

p-value = P( t < -0.43 ) = 0.340061 = 0.340

Conclusion : Since the​ P-value is less than the significance​ level, we 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) A (1-$\alpha$)% confidence interval for population mean difference is obtained as follows :
$(\bar{x}_{D}\mp t_{\alpha /2,n-1}*S_{D}*\sqrt{\frac{1}{n}})$
Thus, 95% confidence interval for population mean difference is :
$(-1546.125-1.8946*10160.37487*\sqrt{\frac{1}{8}},-1546.125+1.8946*10160.37487*\sqrt{\frac{1}{8}})$

$\Rightarrow$ ( -8351.9734, 5259.7234 )
$\Rightarrow$ ( -8352, 5260 )

Since the confidence interval contains zero we fail to reject the null hypothesis.

Hope this answers your query!