Jim is timed 16 times to measure his performance in minutes of a certain task. The average of these 16 observations came out to be 8.5 minutes with a standard deviation of 1.25 minutes. Obtain a 90% confidence interval for the actual mean performance of Jack's work.
Solution :
Given that,
t /2,df = 1.753
Margin of error = E = t/2,df * (s /n)
= 1.753 * (1.25 / 16)
Margin of error = E = 0.55
The 90% confidence interval estimate of the population mean is,
- E < < + E
8.5 - 0.55 < < 8.5 + 0.55
7.95 < < 9.05
(7.95 , 9.05)
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