Answer to Question #236965 in Statistics and Probability for blossomqt

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 cm2. The producer is required to test the bearings before they are shipped, and so the diameters of 16 bearings are measured with a precise instrument, resulting in the following values:

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. Use the data and α = 0.025 to test the data to determine whether the population of these bearings is to be rejected because of too high variance.

Answer 1

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

n=16

sample variance = 0.0014

$H_0: \sigma^2 ≤ 0.001 \\ H_1 : \sigma^2 > 0.001$

Test-statistic:

$χ^2 =\frac{(n-1)s^2}{\sigma^2} \\ χ^2 =\frac{(16-1) 0.001}{0.0014} \\ = 21$

The critical value at α = 0.025 with 15 degrees of freedom is 27.4884

Decision:

Fail to 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.