Question

.11 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

The 99% confidence interval for the mean diameter of pieces from this machine is between 0.93 and 1.07.

Explanation:

Sample mean:

Sum of all values divided by the number of values. So

`M = (1.01+0.97+1.03+1.04+0.99+0.98+0.99+1.01+1.03)/(9) = 1`

Sample standard deviation:

Square root of the sum of the differences between each value and the mean, divided by the 1 less than the sample size. So

`s = \sqrt{((1.01-1)^2+(0.97-1)^2+(1.03-1)^2+(1.04-1)^2+(0.99-1)^2+(0.98-1)^2+(0.99-1)^2+(1.01-1)^2+(1.03-1)^2)/(8)} = 0.0657`

Confidence interval:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 9 - 1 = 8

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.99)/(2) = 0.995`. So we have T = 3.355

The margin of error is:

`M = T(s)/(√(n)) = 3.355(0.0657)/(√(9)) = 0.07`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1 - 0.07 = 0.93

The upper end of the interval is the sample mean added to M. So it is 1 + 0.07 = 1.07

The 99% confidence interval for the mean diameter of pieces from this machine is between 0.93 and 1.07.