Question

A random sample of 100 visitors to a popular theme park spent an average of $142 on the trip with a standard deviation of $47.50. Which of the following would be the 98% confidence interval for the mean amount of money spent by all visitors to this theme park?

a. ($130.77, $153.23)
b. ($132.77, $151.43)
c. ($132.69, $151.31)
d. ($140.88, $143.12)

Answer 1

a) 98% confidence interval for the mean amount of money spent by all visitors to this theme park

($130.77, $153.23)

Explanation:

Explanation:-

Given sample size 'n' = 100

Given the average of $142 on the trip with a standard deviation of $47.50

mean of the sample (x⁻) = $142

standard deviation σ = $47.50.

Confidence intervals:-

The 98% of confidence intervals

`(x^(-) -2.326 (S.D)/(√(n) ),x^(-) +2.326(S.D)/(√(n) ) ))`

`142 -2.326 ((47.50)/(√(100) )),142 +2.326((47.50)/(√(100) ) ))`

on calculation, we get

($130.95 , 153.04)

The confidence intervals are ($130.95 , 153.04)

Final answer:-

Its nearly the given option its (a)

($130.77, $153.23)