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

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 (

Input values above into equation

confidence interval = ( -0.38 , 0.08 )

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all