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?
= (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
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
Sample size = n =221
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