Question

A sample of 105 laramie residents was randomly selected and surveyed about Easter weekend spending. Their 95% confidence interval for the true average amount of money that Laramie residents spend during the Easter weekend was calculated to be $277< μ < $327. National data shows that the standard deviation of money spent duing easter weekend is $91.00.

a.) What is the error?

b.) If I wanted to calculate a new confidence interval and drop my error to $12.00, how large would my sample size have to be?

Answer 1

Solution :

  = (Lower confidence interval + Upper confidence interval ) / 2

Sample mean = = (277+327) / 2=302

Margin of error = E = Upper confidence interval -   = 327-302=25

Margin of error = E =25

(B)

Solution

standard deviation =   =$91.00

Margin of error = E = $12.00

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

sample size = n = [Z/2* / E] 2

n = ( 1.96* 91.00 / 12.00 )2

n =220.9

Sample size = n =221