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3 years ago

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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

1 answer:

Gemiola [76]3 years ago

5 0

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

<u>Determine the 99% confidence interval for difference in proportion</u>

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 )

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Step-by-step explanation:

We can use the following trigonometric identity:

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Solving each equation by b and equaling we have:

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I hope it helps you!

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