Question

A major car dealership has several stores in a big city. The owner wants to determine if there is a difference in the proportion of SUVs that are sold at stores A and B. The owner gathers the sales records for each store from the past year. A random sample of 55 receipts from store A shows that 30 of the sales were for SUVs. Another random sample of 60 receipts from store B shows that 42 of the sales were for SUVs. Assuming conditions for inference are met, what is the 99% confidence interval for the difference in proportions of sales that are SUVs

Answer 1

Answer:

confidence interval = ( -0.38 , 0.08 )

Step-by-step explanation:

Given data :

n1 = 55 , n2 = 60 ,

Let P1 ( proportion of SUV sales at store A ) = 30 / 55 = 0.55

P2 ( sample proportion of SUV sales at store B ) = 42/60 = 0.70

The Z-value at 99% confidence Interval = 2.58

Determine the 99% confidence interval for difference in proportion

applying the formula below

( P1 - P2 ) ± Z ( $\sqrt{p1( 1-p1)/n1 + p2(1-p2)/n2}$

Input values above into equation

confidence interval = ( -0.38 , 0.08 )