1. ## 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)

2. 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 $\displaystyle \mathcal{N}(m,\sigma^2)$, 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.