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Math Help - Estimating the mean using confidence levels

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by angrynapkin View Post
    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 \alpha = 0.1.
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    Is this only when alpha = .1?
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    Quote Originally Posted by angrynapkin View Post
    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 12.2 the standard error of this mean is 10/\sqrt{100}=1, and the sample mean has an approximatly normal distribution.

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

    Do this and you will get your professors answer.

    (Note it is the interval which is the random variable not the population mean)
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