Question

In: Statistics and Probability

According to a​ newspaper, customers are not settling for automobiles straight off the production lines. A...

According to a​ newspaper, customers are not settling for automobiles straight off the production lines. A sample of

169

recent purchases of a particular car model yielded a sample mean of

​$5881

above the​ $20,200 base sticker price. Suppose the cost of accessories purchased for all cars of this model has a standard deviation of

​$1527

a. Calculate an 80​% confidence interval for the average cost of accessories on this model.

b. Determine the margin of error in estimating the average cost of accessories on this model.

c. What sample size would be required to reduce the margin of error by​ 50%?

Solutions

Expert Solution

Solution :

Given that,

a) Z/2 = Z0.10 = 1.282

Margin of error = E = Z/2 * ( /n)

= 1.282 * ( 1527 /  169 )

= 151

At 80% confidence interval estimate of the population mean is,

  ± E

= 5881  ± 151

= ( $ 5730, $ 6032 )

b) Margin of error = E = Z/2 * ( /n)

= 1.282 * ( 1527 /  169 )

= $ 151

c) margin of error = E = 151 / 2 = 75.5

sample size = n = [Z/2* / E] 2

n = [1.282 * 1527 / 75.5 ]2

n = 672.29

Sample size = n = 673

orchestra answered 1 year ago
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