Question 1
Distribution of sample mean
Assume \(x_1,x_2,...,x_n\) is a random sample of size n from a distribution with mean \(\mu\) and standard deviation \(\sigma\).
- Then the sample mean is \(\bar{x}\) = \(\frac{1}{n} \sum_{i=1}^{n}x_i\)
- \(\bar{x}\) follows a distribution with mean \(\mu\) and standard deviation \(\frac{\sigma}{\sqrt{n}}\)
- Sample mean \(\bar{x}\) approximately is \(N(\mu, \frac{\sigma^2}{n})\)
- \(\sigma\) the population standard deviation, which is unknown
- \(\bar{z} = \frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) is approximately 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: \((\bar{y}_1 - \bar{y}_2) \pm t_\frac{\alpha}{2}s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\)
where \(t_\frac{\alpha}{2}\) is the (1-\(\frac{\alpha}{2}\))% quantile of a Student’s t distribution with \(df=n_1 + n_2-2\)