Question

A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.92 years and the standard deviation is 9.73 years a) Construct a 99% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence interval have been met. b) How large is the margin of error? c) How would the confidence interval change if you had assumed that the standard deviation was known to be 10.0 years?

c) Also, would the confidence interval become more narrow, or more wider?

Answer 1

a). The formula for estimation of confidence interval is:

μ = M ± t(s_M)

where:

M = sample mean
t = t statistic determined by confidence level
s_M = standard error = √(s²/n)

Assumption:

1. The Distribution is normal

2. Since the sample size,n =25<30 hence it follows t distribution

3. The sample is randomly selected.

Calculation

M = 32.92
t = 2.8
s_M = √(9.73²/25) = 1.95

μ = M ± t(s_M)
μ = 32.92 ± 2.8*1.95
μ = 32.92 ± 5.4428

CI [27.4772, 38.3628].

We can be 99% confident that the population mean (μ) falls between 27.4772 and 38.3628.

b). The margin of error is calculated as

t(s_M)= 2.8*1.95

=5.4428

c).if the standard deviation changed to 10 years than,

from the above-used formula

s_M = √(10²/25) = 2

μ = M ± t(s_M)
μ = 32.92 ± 2.8*2
μ = 32.92 ± 5.5939

CI [27.3261, 38.5139].

d). The confidence interval becomes wider as we increases the standard deviation.