In: Statistics and Probability
Construct a 95% confidence interval to estimate the population mean when Mean=125 and s = 26 for the sample sizes below.
a)N=40
b)N=70
c) N=100
A.)The 95% confidence interval for the population mean when N=40is from a lower limit of_____to an upper limit of ______.
B.) The 95% confidence interval for the population mean when N=70is from a lower limit of _____to an upper limit of ______.
C.) The 95% confidence interval for the population mean when N=100is from a lower limit of_____to an upper limit of ______.
Solution :
Given that,
= 125
s = 26
(A)
n = 40
Degrees of freedom = df = n - 1 = 40 - 1 = 39
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,39 = 2.023
Margin of error = E = t/2,df * (s /n)
= 2.023 * (26 / 40)
= 8.3
The 95% confidence interval estimate of the population mean is,
- E < < + E
125 - 8.3 < < 125 + 8.3
116.7 < < 133.3
The 95% confidence interval for the population mean when N = 40 is from
a lower limit of 116.7 to an upper limit of 133.3 .
(B)
n = 70
Degrees of freedom = df = n - 1 = 70 - 1 = 69
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,69 = 1.995
Margin of error = E = t/2,df * (s /n)
= 1.995 * (26 / 70)
= 6.2
The 95% confidence interval estimate of the population mean is,
- E < < + E
125 - 6.2 < < 125 + 6.2
118.8 < < 131.2
The 95% confidence interval for the population mean when N = 70 is from
a lower limit of 118.8 to an upper limit of 131.2
(C)
n = 100
Degrees of freedom = df = n - 1 = 100 - 1 = 99
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,99 = 1.984
Margin of error = E = t/2,df * (s /n)
= 1.984 * (26 / 100)
= 5.2
The 95% confidence interval estimate of the population mean is,
- E < < + E
125 - 5.2 < < 125 + 5.2
119.8 < < 130.2
The 95% confidence interval for the population mean when N = 100 is from
a lower limit of 119.8 to an upper limit of 130.2