Confidence Interval

**magentarita** · December 25th 2008, 04:35 AM

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)

**Laurent** · December 25th 2008, 08:18 AM

Originally Posted by magentarita

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

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