1. ## Random sample

A simple random sample of size 10 drawn from a normal distribution yielded a sample mean of 100 and a sample variance of 64.

i) Obtain a 99% confidence interval for the true mean of the population
ii) What would the interval be if the population variance were known to be
σ^2 = 81?
iii) If it is required to be 95% confident that the sample mean is to be correct to within 2 euro of the true mean ,what samle size is necessary? Use σ^2 = 81.

2. Originally Posted by matty888
A simple random sample of size 10 drawn from a normal distribution yielded a sample mean of 100 and a sample variance of 64.

i) Obtain a 99% confidence interval for the true mean of the population
ii) What would the interval be if the population variance were known to be
σ^2 = 81?
iii) If it is required to be 95% confident that the sample mean is to be correct to within 2 euro of the true mean ,what samle size is necessary? Use σ^2 = 81.
I'll assume that you're familiar with the notation I use.

i) $\displaystyle \left( \bar{x}-\frac{t_{\alpha/2} \, s}{\sqrt{n}}, ~ \bar{x} + \frac{t_{\alpha/2} \, s}{\sqrt{n}} \right)$

where t follows a t-distribution with n - 1 degrees of freedom and $\displaystyle t_{\alpha/2}$ is the critical value of t. In your case n = 10 and $\displaystyle \alpha = 0.01 \Rightarrow t_{0.005} = ........$

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ii) $\displaystyle \left( \bar{x}-\frac{z_{\alpha/2} \, \sigma}{\sqrt{n}}, ~ \bar{x} + \frac{z_{\alpha/2} \, \sigma}{\sqrt{n}} \right)$

where z follows a standard normal distribution and $\displaystyle z_{\alpha/2}$ is the critical value of z. In your case n = 10 and $\displaystyle \alpha = 0.01 \Rightarrow z_{0.005} = ........$

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iii) Solve $\displaystyle 2 = \frac{z_{\alpha/2} \, \sigma}{\sqrt{n}}$ for n, where $\displaystyle \alpha = 0.05 \Rightarrow z_{0.025} = ........$