Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. Suppose that the mean income is found to be $18.5 for a random sample of 2253 people. Assume the population standard deviation is known to be $6.1. Construct the 98% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

Answer 1

= ( $18.2, $18.8)

Therefore, the 98% confidence interval (a,b) = ( $18.2, $18.8)

Explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $18.50

Standard deviation r = $6.10

Number of samples n = 2253

Confidence interval = 98%

z(at 98% confidence) = 2.33

Substituting the values we have;

$18.5+/-2.33($6.1/√2253 )

$18.5+/-2.33($0.128513644290)

$18.5+/-$0.299436791196

$18.5+/-$0.3

= ( $18.2, $18.8)

Therefore at 98% confidence interval (a,b) = ( $18.2, $18.8)