Question

A machine produces metal pieces that are cylindrical in shape. A sample of pieces is taken, and the diameters are found to be 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, and 1.03 centimeters. Find a 99% confidence interval for the mean diameter of pieces from this machine, assuming an approximately normal distribution.

Answer 1

Answer:

99% confidence interval: (0.97816,1.03184)          

Step-by-step explanation:

We are given the following data set:

1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, 1.03

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$  

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.  

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{9.05}{9} = 1.005$

Sum of squares of differences = 0.00001975308642 + 0.001264197531 + 0.000597530864 + 0.001186419753 + 0.0002419753088, + 0.00065308642 + 0.0002419753088 + 0.00001975308642 + 0.000597530864 = 0.0048

$S.D = \sqrt{\frac{0.0048}{8}} = 0.024$

Confidence interval:

$\bar{x} \pm t_{critical}\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 8}~\alpha_{0.01} = \pm 3.355$

$1.005 \pm 3.355(\frac{0.024}{\sqrt{9}} ) = 1.005 \pm 0.02684 = (0.97816,1.03184)$