Question

A machine produces cylindrical metal pieces. A sample of pieces is taken, and the diameter are found to be 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01 and 1.03 cm. Find the t-value for a 98% confidence interval for the diameters of the cylindrical metal pieces. Assume that the diameters are approximately normally distributed.

Answer 1

Since the population value of σ is unknown, we want to use the confidence interval for σ unknown. Thus the 1 − α confidence interval is

$\bar{x}-t_{\alpha/2}\cfrac{s}{\sqrt{n}}<\mu<\bar{x}+t_{\alpha/2}\cfrac{s}{\sqrt{n}}$

$\bar{x}-t_{\alpha/2}\cfrac{s}{\sqrt{n}}<\mu<\bar{x}+t_{\alpha/2}\cfrac{s}{\sqrt{n}}$$\bar{x}-t_{\alpha/2}\cfrac{s}{\sqrt{n}}<\mu<\bar{x}+t_{\alpha/2}\cfrac{s}{\sqrt{n}}$

where $t_{\alpha/2}$ is the value from the t-distribution table for

$P(T>t_{\alpha/2})=\alpha/2.$

For a 98% confidence interval we have α = 0.02 and α/2 = 0.01.

Using the table for df = n - 1 = 9 - 1 = 8 (degrees of freedom, n = 9 - the sample size):

$t_{0.01;8}=2.896.$

The population sample mean:

$\bar{x}=(1.01+0.97+1.03+1.04+0.99+0.98+\\+0.99+1.01+1.03)/9=1.0056.$

The population variance:

$s^2=\cfrac{\sum(x_i-\bar x)^2}{n-1}=\\ =((1.01-1.0056)^2+(0.97-1.0056)^2+\\ +(1.03-1.0056)^2+(1.04-1.0056)^2+\\ +(0.99-1.0056)^2+(0.98-1.0056)^2+\\ +(0.99-1.0056)^2+(1.01-1.0056)^2+\\ +(1.03-1.0056)^2)/8=0.00060278.$

The population standard deviation:

$s=\sqrt{0,00060278}=0.0246.$

$s=\sqrt{0,00060278}=0.0246.$

Thus the 98% CI is

$1.0056-2.896\cdot\cfrac{0.0246}{\sqrt{9}}<\mu<1.0056+2.896\cdot\cfrac{0.0246}{\sqrt{9}}\\ 0.9819<\mu<1.0293.$

$1.0056-2.896\cdot\cfrac{0.0246}{\sqrt{9}}<\mu<1.0056+2.896\cdot\cfrac{0.0246}{\sqrt{9}}\\ 0.9819<\mu<1.0293.$$1.0056-2.896\cdot\cfrac{0.0246}{\sqrt{9}}<\mu<1.0056+2.896\cdot\cfrac{0.0246}{\sqrt{9}}\\ 0.9819<\mu<1.0293.$