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Thread: Confidence Interval

  1. #1
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    Confidence Interval

    A math expert kept track of how many questions he answered over one week: 10, 20, 30, 20, 20, 70, 40 What is the 95% confidence interval for the mean number of questions answered each day? (to the nearest question)

    a-(10, 70)
    b-(12, 48)
    c-(26, 34)
    d-(15, 45)
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  2. #2
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    Quote Originally Posted by magentarita View Post
    A math expert kept track of how many questions he answered over one week: 10, 20, 30, 20, 20, 70, 40 What is the 95% confidence interval for the mean number of questions answered each day? (to the nearest question)

    a-(10, 70)
    b-(12, 48)
    c-(26, 34)
    d-(15, 45)
    Assuming that the number of questions answered is a gaussian random variable $\displaystyle \mathcal{N}(m,\sigma^2)$, you have to use the confidence interval for the Student distribution: you know that $\displaystyle T=\sqrt{n}\frac{\overline{X}-m}{\overline{\sigma}}$ is distributed according to a Student distribution with $\displaystyle n-1=7-1=6$ degrees of freedom, where $\displaystyle n$ is the size of the sample, and $\displaystyle \overline{X}$ and $\displaystyle \overline{\sigma}$ are the empirical mean and standard deviation.
    By looking up in tables (or such as), you find a confidence interval for $\displaystyle T$, i.e. a number $\displaystyle t$ such that $\displaystyle P(-t<T<t)=0.95$ or equivalently $\displaystyle P(T>t)=0.025$. This should be $\displaystyle t\simeq 2.45$.
    This gives you a confidence interval for $\displaystyle m$: $\displaystyle P(\overline{X}-\frac{t\overline{\sigma}}{\sqrt{n}}<m<\overline{X} +\frac{t\overline{\sigma}}{\sqrt{n}})=0.95$.
    All that remains is to compute $\displaystyle \overline{X}$, $\displaystyle \overline{\sigma}$ from the data to deduce the bounds of this interval.
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