Estimating the mean using confidence levels

• Jul 14th 2009, 09:55 PM
angrynapkin
Estimating the mean using confidence levels
A company is interested in estimating the mean number of sick days taken by its employees. The firm's statistician selects 100 files at random and ntoes the number of sick days taken by each employee. The mean was 12.2 days and the standard deviation was 10 days. Estimate the mean with 90% confidence and interpret.

I got (11.662, 12.738) with an E of .538 after doing the formula for E

My professor wrote down the answer of (10.555,13.845), which corresponds with an E of 1.645 if you do the subtractions.

1.645 is also the Critical Value for a 90% confidence level, what in the world is going on here?
• Jul 14th 2009, 10:18 PM
mr fantastic
Quote:

Originally Posted by angrynapkin
A company is interested in estimating the mean number of sick days taken by its employees. The firm's statistician selects 100 files at random and ntoes the number of sick days taken by each employee. The mean was 12.2 days and the standard deviation was 10 days. Estimate the mean with 90% confidence and interpret.

I got (11.662, 12.738) with an E of .538 after doing the formula for E

My professor wrote down the answer of (10.555,13.845), which corresponds with an E of 1.645 if you do the subtractions.

1.645 is also the Critical Value for a 90% confidence level, what in the world is going on here?

Confidence intervals are 'two-sided' so you use the two-sided critical value of z when $\displaystyle \alpha = 0.1$.
• Jul 14th 2009, 10:21 PM
angrynapkin
Is this only when alpha = .1?
• Jul 14th 2009, 11:26 PM
CaptainBlack
Quote:

Originally Posted by angrynapkin
A company is interested in estimating the mean number of sick days taken by its employees. The firm's statistician selects 100 files at random and ntoes the number of sick days taken by each employee. The mean was 12.2 days and the standard deviation was 10 days. Estimate the mean with 90% confidence and interpret.

I got (11.662, 12.738) with an E of .538 after doing the formula for E

My professor wrote down the answer of (10.555,13.845), which corresponds with an E of 1.645 if you do the subtractions.

1.645 is also the Critical Value for a 90% confidence level, what in the world is going on here?

The sample mean is $\displaystyle 12.2$ the standard error of this mean is $\displaystyle 10/\sqrt{100}=1$, and the sample mean has an approximatly normal distribution.

Hence the interval $\displaystyle [12.2-\lambda, 12.2+\lambda]$ contains the mean with probability $\displaystyle p$, where $\displaystyle \lambda=P^{-1}(p/2)$ where $\displaystyle P$ denotes the cumulative normal distribution.