A local economist is interested in studying the incomes of consumers in her city. The population...

A local economist is interested in studying the incomes of consumers in her city. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in a mean income of $15,000. What is the width of the 90% confidence interval?

Select one:

a. $364.30

b. $728.60

c. $465.23

d. None of the suggested answers are correct.

e. $232.60

Solutions

Expert Solution

Solution :

Given that,

Population standard deviation = = $1000
Sample size = n =50

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2 * ( / $\sqrt$ n)

= 1.645 * (1000 / $\sqrt$ 50 )

=232.64*2

width=465.23

orchestra answered 1 year ago

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