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Interval estimation

Activity 7.1   Justify the last statement.
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Answer 7.1   We just have to rework the inequality. Going step by step:

$\displaystyle P\left[\left\vert\frac{\bar{X}-\mu}{1/\sqrt{n}}\right\vert\le 1.96\right]$ $\displaystyle =0.95$    
$\displaystyle P\left[-1.96/\sqrt{n}\le \bar{X}-\mu\le 1.96/\sqrt{n}\right]$ $\displaystyle =0.95$    
$\displaystyle P\left[-1.96/\sqrt{n}-\bar{X}\le -\mu\le -\bar{X} +1.96/\sqrt{n}\right]$ $\displaystyle =0.95$    
$\displaystyle P\left[1.96/\sqrt{n}+\bar{X}\ge \mu\ge \bar{X} -1.96/\sqrt{n}\right]$ $\displaystyle =0.95$    
$\displaystyle P\left[\bar{X}-1.96/\sqrt{n}\le \mu\le \bar{X} +1.96/\sqrt{n}\right]$ $\displaystyle =0.95$    


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Activity 7.2   Why don't we always choose a very high confidence for the interval?
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Answer 7.2   We don't always want to use a very high confidence, because the interval would be very long. We have to trade off the length of the interval against the confidence percentage.
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Activity 7.3   Prove the last statement.
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Answer 7.3   We know that $ Z=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$ has a standard normal distribution, and that $ W=S^2/\sigma^2$ has an independent $ \chi^2$ distribution with $ (n-1)$ degrees of freedom. It follows that

$\displaystyle \frac{Z}{\sqrt{W/(n-1)}}=\frac{\bar{X}-\mu}{S/\sqrt{n}}
$

has a Student's t distribution with $ (n-1)$ degrees of freedom.
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Activity 7.4   The results in (7.10) work only for large $ n$, and $ \pi$ not too close to 0 or to 1. If $ n$ is very small then it becomes more difficult to obtain intervals with approximately a 95% confidence. Show that if $ n=1$, and we use the confidence interval $ (0.1,1)$ when there is a success, and $ (0,0.9)$ when there is a failure, we attain confidence of at least 90%, though the actual confidence percentage achieved varies with the true value $ \pi$.
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Answer 7.4   If $ \pi$ is between 0.1 and 0.9, our interval always covers $ \pi$, so for these values the probability of coverage is 100%. This is certainly greater than 90%. If $ \pi < 0.1$ then the probability of covering $ \pi$ is the probability that there is a failure, which is $ 1-\pi$, which is therefore greater than or equal to 90%. Similarly, if $ \pi >0.9$, the probability of covering $ \pi$ is the probability that there is a success, which is $ \pi$, which is greater than 90%.
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next up previous
Next: Hypothesis testing Up: selftestnew Previous: Point estimation
M.Knott 2002-09-12