Question

using T-Score (using Table)

Given the following observations in a simple random sample from a population that is approximately normally distributed , construct and interpret 95% and 99% confidence intervals for the mean.

	Observation
1	66
2	34
3	59
4	56
5	51
6	45
7	38
8	58
9	52
10	52
11	50
12	34
13	42
14	61
15	53
16	48
17	57
18	47
19	50
20	54

Answer 1

Solution:

95% Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 50.35

S = 8.573367575

n = 20

df = n – 1 = 19

Confidence level = 95%

Critical t value = 2.0930

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 50.35 ± 2.0930*8.573367575/sqrt(20)

Confidence interval = 50.35 ± 4.0125

Lower limit = 50.35 - 4.0125 = 46.34

Upper limit = 50.35 + 4.0125 = 54.36

Confidence interval = (46.34, 54.36)

99% Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 50.35

S = 8.573367575

n = 20

df = n – 1 = 19

Confidence level = 99%

Critical t value = 2.8609

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 50.35 ± 2.8609*8.573367575/sqrt(20)

Confidence interval = 50.35 ± 5.4846

Lower limit = 50.35 - 5.4846 = 44.87

Upper limit = 50.35 + 5.4846 = 55.83

Confidence interval = (44.87, 55.83)