Question

A movie theater manager wants to determine whether popcorn sales have increased since the theater switched from using "butter-flavored topping" to real butter. Historically the average popcorn revenue per weekend day was approximately $3,500. After the theater started using real butter, the manager randomly sampled 12 weekend days and calculated the sample’s summary statistics. The average revenue per weekend day in the sample was approximately $4,200 with a standard deviation of $140. Select the function that would correctly calculate the 90% range of likely sample means.

Answer 1

90% confidence intervals for the mean

( 4111.05 , 4288.95)

Explanation:

Step(i):-

Given sample size 'n' = 12

Mean of the sample 'x⁻' = 4200

Standard deviation of the sample 's' = 140

Step(ii):-

90% confidence intervals for the mean is determined by

`(x^(-) - t_{(\alpha )/(2) } (S)/(√(n) ) , x^(-) + t_{(\alpha )/(2) } (S)/(√(n) ))`

`(x^(-) - t_{(0.10)/(2) } (S)/(√(n) ) , x^(-) + t_{(0.10)/(2) } (S)/(√(n) ))`

`(x^(-) - t_(0.05 ) (S)/(√(n) ) , x^(-) + t_(0.05) (S)/(√(n) ))`

The degrees of freedom

ν = n-1 = 12-1 =11

t₀.₀₅ = 2.201

Step(iii):-

`(4200 - 2.201(140)/(√(12) ) ,(4200 -+2.201(140)/(√(12) ))`

( 4200 - 88.95 , 4200+88.95)

( 4111.05 , 4288.95)

Conclusion:-

90% confidence intervals for the mean

( 4111.05 , 4288.95)