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Chapter 11: Problem 70

A company manufactures ball bearings that are supplied to other companies. The machine that is used to manufacture these ball bearings produces them with a variance of diameters of \(.025\) square millimeter or less. The quality control officer takes a sample of such ball bearings quite often and checks, using confidence intervals and tests of hypotheses, whether or not the variance of these bearings is within \(.025\) square millimeter. If it is not, the machine is stopped and adjusted. A recently taken random sample of 23 ball bearings gave a variance of the diameters equal to \(.034\) square millimeter. a. Using a \(5 \%\) significance level, can you conclude that the machine needs an adjustment? Assume that the diameters of all ball bearings have a normal distribution. b. Construct a \(95 \%\) confidence interval for the population variance.

Short Answer

Expert verified
a. You cannot conclude that the machine needs an adjustment because the calculated chi-square value (30.32) is less than the chi-square critical value (35.17) at a 5% significance level, hence, we fail to reject the null hypothesis. b. The 95% confidence interval for the population variance is [0.019, 0.068] square millimeter.

Step by step solution

01

Define null and alternative hypothesis

In order to conduct a hypothesis test, the null and alternative hypotheses must be identified. The null hypothesis \(H_0\) is that the variance is equal to \(.025\) and the alternative hypothesis \(H_1\) suggests that the variance is greater than \(.025\). The hypotheses can be stated as follows: \\(H_0: \sigma^2 = 0.025\) \\(H_1: \sigma^2 > 0.025\)
02

Compute Chi-Square Statistic

The chi-square statistic can be computed using the formula: \\(\chi^2 = (n - 1)s^2 / \sigma^2\) \where \(n\) is the sample size, \(s^2\) is the sample variance, and \(\sigma^2\) is the population variance under \(H_0\). Substituting the given values, we have \\(\chi^2 = (23 - 1)*0.034 / 0.025 = 30.32\)
03

Determine the critical value and make the decision

The critical value at 5% significance level and 22 degrees of freedom (df = n - 1) from the chi-square distribution table is approximately 35.17. If the calculated chi-square value is greater than the critical value, the null hypothesis is rejected. In this case, 30.32 is less than 35.17, therefore, the null hypothesis is not rejected. This suggests that there is not enough statistical evidence at the 5% level of significance to conclude that the machine needs adjustment.
04

Compute the 95% confidence interval for the population variance

The 95% confidence interval for the population variance can be calculated by using the formula: \\([ (n-1)s^2 / \chi^2_{1- \alpha/2, n-1}, (n-1)s^2 / \chi^2_{\alpha/2, n-1}]\) \Where \(\chi^2_{\alpha/2, n-1}\) is the chi-square value for alpha/2 with n-1 degrees of freedom. Given the values n = 23, s虏 = 0.034 and the chi-square values \(\chi^2_{0.025,22} = 10.982\) and \(\chi^2_{0.975,22} = 38.076\), the confidence interval for the variance is \\([ (22)*0.034 / 38.076, (22)*0.034 / 10.982 ] = [0.019, 0.068]\)

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Most popular questions from this chapter

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