2 pts Starting salaries of 64 college graduates who have tak

(2 pts) Starting salaries of 64 college graduates who have taken a statistics course have a mean of $44,500 with a standard deviation of $6,800. Find an 80% confidence interval for . (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.) Lower-bound: Upper-bound:

Solution

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 =    0.1          
X = sample mean =    44500          
z(alpha/2) = critical z for the confidence interval =    1.281551566          
s = sample standard deviation =    6800          
n = sample size =    64          
              
Thus,              
Margin of Error E =    1089.318831          
Lower bound =    43410.68117          
Upper bound =    45589.31883          
              
Thus, the confidence interval is              
              
(   43410.68117   ,   45589.31883   ) [ANSWER]

(2 pts) Starting salaries of 64 college graduates who have taken a statistics course have a mean of $44,500 with a standard deviation of $6,800. Find an 80% con

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