Question

A random sample of eight drugstores shows the following prices (in $ of a popular pain reliever:

3.50, 4.00, 2.00, 3.00, 2.50, 3.50, 2.50, 3.00

Assume the normal distribution for the underlying population to construct the 90% confidence interval for the population mean.

Answer 1

$\text{The confidence interval for the mean }\\ \text{can be calculated as follows:}\\ \bar x±t_{(\frac{α}{2},n-1)}(\frac{s}{\sqrt{n}}),\\ \bar x=\frac{3.5+4+...+3}{8}=3\\ t_{(\frac{α}{2},n-1)}=t_{(0.05,7)}=1.895,\\ s^2=\frac{(3.5-3)^2+(4-3)^2+...+(3-3)^2}{7}=\frac{3}{7}\\ s=\sqrt{s^2}\approx0.655 \\n=8\\ \text{ the lower limit }= 3-1.895(\frac{0.655}{\sqrt{8}})\\ ≈2.56,\\ \text{ the upper limit} = 3+1.895(\frac{0.655}{\sqrt{8}})\\ ≈3.44,\\ \text{so the confidence interval is}\\ 2.56\leq \bar x \leq 3.44\\$