Answer :
Answer:
Step-by-step explanation:
Corresponding means for population 1 and population 2 form matched pairs.
The data for the test are the differences between the mean for population 1 and mean for population 2.
μd = the mean for population 1 minus the mean for population 2.
Population 1 population 2 diff
19 24 - 5
25 27 - 2
31 36 - 5
52 53 - 1
49 55 - 6
34 34 0
59 66 - 7
47 51 - 4
17 20 - 3
51 55 - 4
Sample mean, xd
= (- 5 - 2 - 5 - 1 - 6 + 0 - 7 - 4 - 3 - 4)/10 = - 3.7
xd = - 3.7
Standard deviation = √(summation(x - mean)²/n
n = 10
Summation(x - mean)² = (- 5 + 3.7)^2 + (- - 2 + 3.7)^2 + (- 5 + 3.7)^2+ (- 1 + 3.7)^2 + (- 6 + 3.7)^2 + (0 + 3.7)^2 + (- 7 + 3.7)^2 + (- 4 + 3.7)^2 + (- 3 + 3.7)^2 + (- 4 + 3.7)^2 = 73.7
Standard - eviation = √(73.7/10
sd = 2.71
For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 10 - 1 = 9
The formula for determining the test statistic is
t = (xd - μd)/(sd/√n)
t = (- 3.7 - 0)/(2.71/√10)
t = - 4.32
We would determine the probability value by using the t test calculator.
p = 0.00097
Since alpha, 0.1 > than the p value, 0.00097, then we would reject the null hypothesis. Therefore, at 0.1 level of significance, we can conclude that these data are sufficient to indicate that the mean for population 2 is larger than that for population 1.
3) for population 1,
Mean = (19 + 25 + 31 + 52 + 55 + 34 + 59 + 47 + 17 + 51)/10 = 38.4
Summation(x - mean)² = (19 - 38.4)^2 + (25 - 38.4)^2 + (31 - 38.4)^2+ (52 - 38.4)^2 + (49 - 38.4)^2 + (34 - 38.4)^2 + (59 - 38.4)^2 + (47 - 38.4)^2 + (17 - 38.4)^2 + (51 - 38.4)^2 = 2042.4
Standard deviation, s1 = √2042.4/10 = 14.3
for population 2,
Mean = (24 + 27 + 36 + 53 + 55 + 34 + 66 + 51 + 20 + 55)/10 = 42.1
Summation(x - mean)² = (24 - 42.1)^2 + (27 - 42.1)^2 + (36 - 42.1)^2 + (53 - 42.1)^2 + (55 - 42.1)^2 + (34 - 42.1)^2 + (66 - 42.1)^2 + (51 - 42.1)^2 + (20 - 42.1)^2 + (55 - 42.1)^2 = 2248.9
Standard deviation, s2 = √2248.9/10 = 15
The formula for determining the confidence interval for the difference of two population means is expressed as
Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)
For a 90% confidence interval, we would determine the z score from the t distribution table because the number of samples are small
Degree of freedom =
(n1 - 1) + (n2 - 1) = (10 - 1) + (10 - 1) = 18
z = 1.734
x1 - x2 = 38.4 - 42.1 = - 3.7
√(s1²/n1 + s2²/n2) = √(14.3²/10 + 15²/10)
= 6.55
Margin of error = 1.734 × 6.55 = 11.4
The 90% confidence interval is
- 3.7 ± 11.4