12 Male Female 25,790 21,639 14,524 26,552 4,790 6,120 27,190 18,343 23,516 13,0
ID: 2908561 • Letter: 1
Question
12
Male Female
25,790 21,639
14,524 26,552
4,790 6,120
27,190 18,343
23,516 13,075
9,020 15,315
20,395 11,222
16,075 20,046
27,108 12,563
23,240 19,289
13,190 16,179
10,282 14,298
13,281 17,418
22,976 22,907
17,486 4,619
2,146 19,344
20,158 24,087
11,920 20,435
20,802 17,842
16,884 12,486
17,365 28,420
21,765 7,031
7,097 17,476
7,173 4,258
25,954 12,014
10,884 19,090
13,813 13,630
13,248 20,831
12,347 19,269
19,060 14,442
12,243 35,301
7,981 6,795
19,142 11,298
18,846 10,983
8,608 29,266
13,142 20,541
14,738 17,875
12,165 19,656
17,407 18,569
9,915 17,736
12,574 19,907
4,312 11,706
19,002 16,990
15,071 15,201
17,981 27,148
16,169 38,736
20,416 25,597
36,882 34,185
16,831 23,234
45,982 33,077
25,237 20,141
10,856 7,843
15,409 24,376
8,122 15,202
7,254 12,946
21,220 29,444
Explanation / Answer
Part a
The null and alternative hypotheses for this test are given as below:
Null hypothesis: H0: There is no significant difference in the average number of words spoken by the male and female among couples.
Alternative hypothesis: H1: The average number of words spoken by the male is fewer than female among the couples.
H0: µd = 0
H1: µd < 0
From given data, we have
Level of significance = ? = 0.05
Sample size = n = 56
Degrees of freedom = n – 1 = 55
Dbar = -2088.1786
Sd = 8959.8325
Test statistic = t = Dbar / [Sd/sqrt(n)]
t = -2088.1786/( 8959.8325/sqrt(56))
t = -1.74406
t = -1.74
P-value = 0.043
(By using excel)
P-value < ? = 0.05
So, we reject the null hypothesis that there is no significant difference in the average number of words spoken by the male and female among couples.
Conclusion
Since the P-value is less than or equal to the significance level, we reject the null hypothesis. There is sufficient evidence to support the claim that males speak fewer words in a day than females.
Part b
Here, we have to find the confidence interval for the difference between population means.
Confidence interval = Dbar ± t*Sd/sqrt(n)
From above part, we have
Level of significance = ? = 0.05
Sample size = n = 56
Degrees of freedom = n – 1 = 55
Confidence level = 1 – ? = 1 – 0.05 = 0.95 = 95%
Critical t value = 2.0040
Dbar = -2088.1786
Sd = 8959.8325
(by using t-table/excel)
Confidence interval = -2088.1786 ± 2.0040*8959.8325/sqrt(56)
Confidence interval = -2088.1786 ± 2399.405
Lower limit = -2088.1786 - 2399.405 = -4487.5836
Upper limit = -2088.1786 + 2399.405 = 311.2264
The confidence interval is -4487.6 < µd < 311.2
Since the confidence interval contains positive and negative numbers, we fail to reject the null hypothesis.
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