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

Answer:

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

($130.77, $153.23)

Step-by-step 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

$(x^{-} -2.326 \frac{S.D}{\sqrt{n} },x^{-} +2.326\frac{S.D}{\sqrt{n} } ))$

$142 -2.326 (\frac{47.50}{\sqrt{100} }),142 +2.326(\frac{47.50}{\sqrt{100} } ))$

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)