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According to a newspaper, customers are not settling for automobiles straight off the production lines. A sample of 173 recent purchases of a particular car model yielded a sample mean of $4,878 above the $20,200 base sticker price. Suppose the cost of 
accessories purchased for all cars of this model has a standard deviation of $1,568.
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%?

First of, this question does not specify whether to use the t-distribution of the standard normal distribution to construct the confidence interval.  

There are cases where I have seen instructors using the t-distribution whenever the sample size is greater than 30 to construct the confidence interval. The reserved procedure is to always use the standard normal distribution to construct the confidence interval whenever the standard deviation of the sample is not given.  Therefore, the standard normal-distribution is used in this case. 

To begin;- 

Calculate the z-score for the 80% confidence interval using excel. The function;- =NORM.S.INV(1-0.2/2) = 1.282

The 80% confidence interval is then calculated using this formula;- X_bar +/-  z-score * Standard Error

Notice that, we are told that the sample mean is $4,878 above the $20,2000 which means the sample mean will be obtained by adding the two. 

X_bar  = $20,200 + $4,878 = 25078

Confidence interval = $25,078 +/- 1.282 * 1568/sqrt(173)

The margin of error is =  1.282 * 1568/sqrt(173) = 152.831

Therefore, the 80% confidence interval is

$25,078 +/-  152.831  = ($24,925.17, $25,230.83)


The required sample size formula for the construction of a given confidence interval is always calculated using the formula;;-

n = ((Z-score * standard deviation)/Margin of Error)^2

To reduce the margin of error by 50%, we need to multiply the margin of error found by 0.50;- 0.50 * 152.831  =  76.416 (This is the targeted margin of error), and inserting this to the formula above;-

n = ((1.282 * 1568)/76.416)^2  =  692 

Therefore, the sample size that is needed to reduce the calculated margin of error by half is 292








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