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
[tex](x^{-} -2.326 \frac{S.D}{\sqrt{n} },x^{-} +2.326\frac{S.D}{\sqrt{n} } ))[/tex]
[tex]142 -2.326 (\frac{47.50}{\sqrt{100} }),142 +2.326(\frac{47.50}{\sqrt{100} } ))[/tex]
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)
