Since the confidence interval contains (only positive numbers/zero/only negative
ID: 3326147 • Letter: S
Question
Since the confidence interval contains (only positive numbers/zero/only negative numbers) (fail to reject/reject) the null hypothesis. 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 16.305 27,155 1370 7910 19,175 15.30914071 25,357 Female 25,085 13,395 19,026 18,376 13,134 16,336 15,685 19,023 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, 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? Ho:Ha! word(s) Type integers or decimals. Do not round) Identify the test statistic t(Round to two decimal places as needed) Identify the P-value P-value (Round to three decimal places as needed )
Explanation / Answer
Solution:
Part a
Here, we have to use paired t test. Null and alternative hypothesis for this test are given as below:
H0: µd = 0
H1: µd < 0
We are given
= 0.05
Test statistic formula is given as below:
t = Dbar / [Sd/sqrt(n)]
From given data, we have
Sample size = 8
Dbar = -1301.0000
Sd = 10401.0140
df = n – 1 = 8 – 1 = 7
Calculation table is given as below:
Male
Female
Di
(Di - DBar)^2
16305
25085
-8780
55935441
27155
13395
13760
226833721
1370
19026
-17656
267486025
7910
18376
-10466
83997225
19175
13134
6041
53904964
15309
13336
1973
10719076
14071
15685
-1614
97969
25357
19023
6334
58293225
t = -1301.0000/[10401.0140/sqrt(8)]
t = -0.35
P-value = 0.3670
(By using t-table or excel)
Since the P-value is greater than the significance level, we do not reject the null hypothesis. There is not sufficient evidence to support the claim that males speak fewer words in a day than females.
Part b
From given data, we have
Sample size = 8
Dbar = -1301.0000
Sd = 10401.0140
df = n – 1 = 8 – 1 = 7
= 0.05
Critical value = t =
Confidence interval = Dbar -/+ t*Sd/sqrt(n)
Confidence interval = -1301.0000 -/+ 2.3646*10401.0140/sqrt(8)
Confidence interval = -1301.0000 -/+ 2.3646* 3677.313765
Confidence interval = -1301.0000 -/+ 8695.4653
Lower limit = -1301.0000 - 8695.4653 = -9996.5
Upper limit = -1301.0000 + 8695.4653 = 7394.5
Since the confidence interval contains zero, so we fail to reject the null hypothesis.
Male
Female
Di
(Di - DBar)^2
16305
25085
-8780
55935441
27155
13395
13760
226833721
1370
19026
-17656
267486025
7910
18376
-10466
83997225
19175
13134
6041
53904964
15309
13336
1973
10719076
14071
15685
-1614
97969
25357
19023
6334
58293225