Random sample of 87 airline pilots had an average yearly inc

Random sample of 87 airline pilots had an average yearly income of $99,400 with a standard deviation of $12,000. If we want to determine of 95% confidence interval yearly income, what is the value of t, and develop a 95% interval for the average yearly income of all pilots?

Solution

a)
As

df = n - 1 = 87 - 1 = 86,

then for 95% confidence,

t(alpha/2) = critical t for the confidence interval =    1.987934206   [ANSWER, VALUE OF T]

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Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    99400          
t(alpha/2) = critical t for the confidence interval =    1.987934206          
s = sample standard deviation =    12000          
n = sample size =    87          

df = n - 1 =    86          

Thus,              
              
Lower bound =    96842.45298          
Upper bound =    101957.547          
              
Thus, the confidence interval is              
              
(   96842.45298   ,   101957.547   ) [ANSWER, CONFIDENCE INTERVAL]