Question 1.
Distribution of sample mean
Assume \(x_1,x_2,...,x_n\) is a random sample size \(n\) from a distribution with mean \(\mu\) and standard deviation \(\sigma\).
Then the sample mean is \[\bar{x} = \frac{1}{n}\sum\limits_{i=1}^{n} x_i\]
\(\bar{x}\) follows a distribution with mean \(\mu\) and standard deviation \(\sigma/\sqrt{n}\)
Sample mean \(\bar{x}\) approximately is \(\text{N}(\mu,\sigma^2/n)\)
\(\sigma\) the population standard deviation, which is unknown
\(\bar{z}=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\) is approximately \(\text{N}(0,1)\)
Question 2.
The \(100(1-\alpha)\)% confidence interval (CI) for the difference \(\mu_1-\mu_2\)
is constructed as follows:
Difference between sample means \(d=\bar{y}_1 -\bar{y}_2\)
Set \(s_p=\sqrt\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}\)
*CI:\((\bar{y}_1-\bar{y}_2)\pm{t_{\alpha/2}sp\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
where \(t_{\alpha/2}\) is the \((1-\alpha/2)\)% quantile of a Student’s t distribution with \(df=n_1+n_2-2\)