Question

A random sample of 31 residents in a particular city indicates an average annual income of $51,950 with a sample standard deviation of $7,785. Construct a 95% confidence interval to estimate the true average annual income for all residents in the city.

Do not round intermediate calculations. Round your final answers to 2 decimal places. Omit the "$" sign.

Lower bound for confidence interval =    dollars, Upper bound for confidence interval =  dollars.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $ 51,950

sample standard deviation = s = $ 7,785

sample size = n = 31

Degrees of freedom = df = n - 1 = 31 - 1 = 30

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t _{0.025, 30} = 2.042

Margin of error = E = t_/2,df * (s /n)

= 2.042 * ( 7785 / 31)

Margin of error = E = 2855.18

The 95% confidence interval estimate of the population mean is,

  ± E

= 51950 ± 2855.18

= $ 49,094.82, $ 54,805.18

Lower bound = 49,094.82 dollars,

Upper bound = 54,805.18 dollars.