According to a newspaper customers are not settling for auto

According to a newspaper, customers are not settling for automobiles straight off the production lines. A sample of 191 recent purchases of a particular car model yielded a sample mean of 4,437 above the $20,200 base sticker price. Suppose the cost of accessories purchased for all cars of this model has a standard deviation of $1,544. a. Calculate a 90% 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%?

Solution

A)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean = 20200 + 4437 =   24637          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    1544          
n = sample size =    191          
              
Thus,              
Margin of Error E =    183.7629141          
Lower bound =    24453.23709          
Upper bound =    24820.76291          
              
Thus, the confidence interval is              
              
(   24453.23709   ,   24820.76291   )
[ANSWER]

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B)

As in A,

Margin of Error E =    183.7629141   [ANSWER]

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c)

As the margin of error varies inversely as the square root of n, then to reduce E to half, then we have a sample size 4 times larger.

Thus,         

n = 191*4 = 764 [ANSWER]