Question

Male   Female
26,616   19,037
15,642   24,613
4,073   5,930
28,733   18,953
24,660   10,286
7,959   14,542
20,806   10,640
18,184   18,336
27,071   12,203
23,508   17,870
12,408   17,792
12,074   14,180
13,336   15,044
19,876   22,004
20,499   5,758
2,194   19,297
20,312   22,900
11,577   17,897
18,808   16,171
15,449   14,824
18,001   28,902
20,374   7,189
6,471   18,205
8,641   6,102
24,494   10,864
10,659   20,285
14,194   12,742
12,715   21,378
11,844   19,696
17,716   13,700
12,804   33,016
7,986   8,591
19,353   9,995
16,284   12,521
12,507   30,372
16,452   22,523
13,507   17,665
8,976   20,305
19,118   20,910
12,726   14,065
13,673   20,063
5,203   13,468
18,474   17,716
15,123   17,715
15,034   27,440
15,184   36,729
19,431   27,376
38,155   34,509
16,824   23,180
48,891   33,402
24,920   16,967
10,260   7,047
14,121   25,900
10,808   16,384
6,112   13,014
20,730   26,957

I a. Use a 0.05 significance level to test the dlaim that among couples, males speak fewer words in a day than females. In this example, ud 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 hypothes s test? 8B Type integers or decimals. Do not round.) Identify the test statistic. t_Џ(Round to two decimal places as needed ) dentify the P-value· (Round to three dermal places as needed ) What is the conclusion based on the hypothes is test? Since the P-value is 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 s □ word(s)<pd-U word(s). P-value. the significance level, the nul hypothesis. Theresufficent evidence to support the claim that males speak fewer words in a day than females. Round to one decimal place as needed.) What feature of the confidence interval leads to the same conclusion reached in part (a)? Since the confidence interval contains V t the null hypothesis.

Answer 1

First, calculate the difference between each pair of data. Make a new column and subtract females from males. We will be working on this column on data.

Male	Female	Difference
26,616	19037	7579
15,642	24613	-8971
4,073	5930	-1857
28,733	18953	9780
24,660	10286	14374
7,959	14542	-6583
20,806	10640	10166
18,184	18336	-152
27,071	12203	14868
23,508	17870	5638
12,408	17792	-5384
12,074	14180	-2106
13,336	15044	-1708
19,876	22004	-2128
20,499	5758	14741
2,194	19297	-17103
20,312	22900	-2588
11,577	17897	-6320
18,808	16171	2637
15,449	14824	625
18,001	28902	-10901
20,374	7189	13185
6,471	18205	-11734
8,641	6102	2539
24,494	10864	13630
10,659	20285	-9626
14,194	12742	1452
12,715	21378	-8663
11,844	19696	-7852
17,716	13700	4016
12,804	33016	-20212
7,986	8591	-605
19,353	9995	9358
16,284	12521	3763
12,507	30372	-17865
16,452	22523	-6071
13,507	17665	-4158
8,976	20305	-11329
19,118	20910	-1792
12,726	14065	-1339
13,673	20063	-6390
5,203	13468	-8265
18,474	17716	758
15,123	17715	-2592
15,034	27440	-12406
15,184	36729	-21545
19,431	27376	-7945
38,155	34509	3646
16,824	23180	-6356
48,891	33402	15489
24,920	16967	7953
10,260	7047	3213
14,121	25900	-11779
10,808	16384	-5576
6,112	13014	-6902
20,730	26957	-6227

Ho: μd >= 0, The mean words spoken by males is greater than mean words spoken by females in a day

Ha: μd < 0 The mean words spoken by males is fewer than mean words spoken by females in a day

This is a paired t-test. The formula to calculate the t-statistic is:

calc S'

d-bar = -1850.36

s =9110.08

n = 56

Hence, t = 1.52

p-value at df = 55 is 0.067 (One-tailed)

Since the p-value is more than the significance level, we cannot reject the null hypothesis. We do not have enough evidence to support the claim that males speak less words then females in a day.

b) Confidence interval can be calculated by the formula:

Lower bound:Xp - ta/2 * Upper bound:XD T Ca/2 * N) r0.

Lower bound: -1850.36 - 2.004 * 1217.38 = -4290.0

Upper bound: -1850.36 + 2.004 * 1217.38 = 589.3

As the confidence interval contains 0, we cannot reject the null hypothesis.