Question

The President of school wants to know the ages of Math students at school. He wants a range for the ages. Construct a 99% confidence interval for the population mean age.

Ages (years)
31
30
21
20
44
43
23
25
30
28
19
21
21
37
26
20
19
22
28
40
20
48
19
20
36
20
20
19
19
21
47
19
21
19
17
19
20
20
32
23
18
21
22
20
19
35
21
42
20
19

1. Sample Mean
2. Sample Standard Deviation

What is the sample size?

What is the confidence level?

What is the lower bound of the confidence interval?

What is the upper bound of the confidence interval?

What is the error bound margin?

Answer 1

a)
sample mean, xbar = 25.28

b)
sample standard deviation, s = 8.5572

c)
sample size, n = 50

d)
confidence level = 99%

e)
degrees of freedom, df = n - 1 = 49

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.68

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (25.28 - 2.68 * 8.5572/sqrt(50) , 25.28 + 2.68 * 8.5572/sqrt(50))
CI = (22.04 , 28.52)

Lower bound = 22.04

upper bound = 28.52

Margin of error,
ME = tc * s/sqrt(n)
ME = 2.68 * 8.5572/sqrt(50)
ME = 3.24