SOLUTION: Nine software engineers were interviewed to develop an estimate for software lines of code. The mean estimate is 12480 lines of code with a standard deviation of 2400 lines of code

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Question 459888: Nine software engineers were interviewed to develop an estimate for software lines of code. The mean estimate is 12480 lines of code with a standard deviation of 2400 lines of code. Develop an 80% confidence interval for the population mean. (Carry intermediate calculations to three decimal places.)



11362 ≤ µ ≤ 13598
12107 ≤ µ ≤ 12853
11769 ≤ µ ≤ 13191
11374 ≤ µ ≤ 13586

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
xbar - t(s/sqrt(n)) ≤ µ ≤ xbar + t(s/sqrt(n))


12480 - 1.397(2400/sqrt(9)) ≤ µ ≤ 12480 + 1.397(2400/sqrt(9))


Note: Use a table or a calculator to find the t-critical value when the confidence level is 80% and the degrees of freedom = 9-1 = 8. This value is t = 1.397


12480 - 1.397(2400/3) ≤ µ ≤ 12480 + 1.397(2400/3)


12480 - 1.397(800) ≤ µ ≤ 12480 + 1.397(800)


12480 - 1117.6 ≤ µ ≤ 12480 + 1117.6


11362.4 ≤ µ ≤ 13597.6


11362 ≤ µ ≤ 13598 (note: I'm rounding to the nearest whole number)


So the 80% confidence interval is 11362 ≤ µ ≤ 13598


So the answer is the first choice.