An automobile manufacturer claims that the mean number of tr

An automobile manufacturer claims that the mean number of trouble-free miles given by a new tire introduced by the company is more than 36,000 miles. When a random sample of sixteen tires was tested, it gave a sample mean of 39,000 miles with a sample standard deviation of 8000 miles. Assume that the lifetime of tires follow a normal distribute Compute a 95percentage confidence interval for the true mean lifetime of a tire. What is the standard error

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 =    0.025          
X = sample mean =    39000          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    8000          
n = sample size =    16          
              
Thus,              
Margin of Error E =    3919.927969          
Lower bound =    35080.07203          
Upper bound =    42919.92797          
              
Thus, the confidence interval is              
              
(   35080.07203   ,   42919.92797   ) [ANSWER]

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b)

SE = s/sqrt(n) = 8000/sqrt(16) = 2000 [ANSWER]