Answer to Question #240679 in Statistics and Probability for Marie

Question

A manufacturing company produces bearings. One line of bearings is specified to be 1.64 centimeters (cm) in diameter. A major customer requires that the variance of the bearings be no more than 0.001 cm. squared. The producer is required to test the bearings before they are shipped, and so the diameters of 16 bearings are measured with a precise measurement. The resulting values are as follows 1.69, 1.62, 1.63, 1.70, 1.66, 1.63, 1.65, 1.71, 1.64, 1.69, 1.57, 1.64, 1.59, 1.66, 1.63, 1.65 Assume bearing diameters are normally distributed and αlpha (a) = 0.025, determine whether the population of these bearings is to be rejected because of too high variance.

Answer 1

SOLUTION

We have given the claim that the variance of the bearings be no more than 0.001 cm²

From given data, summary statistic are

Using R

> var(data)
[1] 0.001446667
> data = c( 1.69, 1.62, 1.63, 1.70, 1.66, 1.63, 1.65, 1.71, 1.64, 1.69, 1.57, 1.64, 1.59, 1.66, 1.63, 1.65)
> length(data)
[1] 16
> var(data)
[1] 0.001446667

$n=16$

sample variance $=0.0014$

$H_0:\sigma^2\le0.001$

$H_a:\sigma^2\gt0.001$

Test-statistics:

$X^2=\frac{(n-1)s^2}{\sigma^2}$

$X^2=\frac{(16-1)0.001}{0.0014}$

$X^2=21$

The critical value at $\alpha=0.025$ with $15$ degrees of freedom is $27.4884$

Decision:

Do not Reject null hypothesis because test statistic value $(21)$ is less than critical value $27.4884$

Conclusion:

There is not sufficient evidence to the population of these bearings is to be rejected because of too high variance.