Standard error = .385/sqrt ((44)) = 0.058
E = 1.96 *. 385 sqrt(44) = 5.00
Mean - E, Mean + E
126-5, 126 + 5
($121, $131)
Standard error = .385/sqrt ((22) = 0.082
E = 1.96 * .385 sqrt(22) = 3.5
Mean - E, Mean + E
126- 3.5, 126 + 3.5
( $122.5, $129.5)
Hoping to lure more shoppers downtown, a city builds a new parking garage in the central business district. The city hopes that the parking revenues will pay for the garage.
During a recent two month period (44 days), the city statistician calculated the 95% confidence interval for the mean daily parking fees collected as:
($121, $131)
If the sample used 22 days rather than 44 days, how would you expect the 95% confidence interval to change? Support your answer.
A) The confidence interval would be narrower since the standard error would be smaller due to the smaller sample size.
B) The confidence interval would be very similar since we are still estimating parking rate income.
C) Given the confidence level of 95%, the interval calculated using a sample of 22 would be about the same since the confidence is not very high.
D) The confidence interval would be wider since the standard error would be larger due to the smaller sample size.
1 answer