Question

A producer of machine parts claimed that the diameters of the connector rods produced by his plant had a variance of at most .03 inch \(^{2}\). A random sample of 15 connector rods from his plant produced a sample mean and variance of .55 inch and .053 inch \(^{2}\), respectively. a. Is there sufficient evidence to reject his claim at the \(\alpha=.05\) level of significance? b. Find a \(95 \%\) confidence interval for the variance of the rod diameters.

Short answer

Also, what is the 95% confidence interval for the variance of the rod diameters? Yes, there is enough evidence to reject the producer's claim, as the test statistic (27.6) is greater than the critical value (23.68). The 95% confidence interval for the variance of the rod diameters is (0.0309, 0.203) inch².

Step-by-step solution

State the null and alternative hypotheses

The null hypothesis (H\(_0\)) is that the population variance is equal to the producer's claim (0.03 inch\(^2\)). The alternative hypothesis (H\(_1\)) is that the population variance is greater than the claimed variance. H\(_0\): \(\sigma^2 = 0.03\) H\(_1\): \(\sigma^2 > 0.03\)

Calculate the test statistic

We will use the chi-square test statistic for this hypothesis test. It is calculated as follows: \(\chi^2 = \frac{(n - 1)s^2}{\sigma_0^2}\) where n is the sample size, s\(^2\) is the sample variance, and \(\sigma_0^2\) is the population variance under the null hypothesis. \(\chi^2 = \frac{(15 - 1)(0.053)}{0.03} = 27.6\)

Find the critical value

To find the critical value, we look up the chi-square distribution table for the given significance level (\(\alpha = 0.05\)) and degrees of freedom (n - 1). The degrees of freedom are: df = n - 1 = 15 - 1 = 14. Looking at a chi-square table, we find the critical value for a one-tailed test (since our alternative hypothesis is \(\sigma^2 > 0.03\)) to be \(\chi^2_{0.05, 14} = 23.68\).

Compare the test statistic to the critical value

Since our test statistic (27.6) is greater than the critical value (23.68), we reject the null hypothesis. There is sufficient evidence at the 5% significance level to reject the producer's claim that the variance of the diameters is at most 0.03 inch\(^2\).

Find the 95% confidence interval for the variance

For a 95% confidence interval, we find the chi-square values that correspond to the lower and upper tails of the distribution: \(\chi^2_\text{upper} = \chi^2_{0.025, 14} = 3.94\) \(\chi^2_\text{lower} = \chi^2_{0.975, 14} = 26.12\) Now we calculate the confidence interval using the following formulas: Lower limit: \(\frac{(n-1) s^2}{\chi^2_\text{upper}}\) Upper limit: \(\frac{(n-1) s^2}{\chi^2_\text{lower}}\) Lower limit: \(\frac{(15-1)(0.053)}{26.12} = 0.0309\) Upper limit: \(\frac{(15-1)(0.053)}{3.94} = 0.203\) Therefore, the 95% confidence interval for the variance of the rod diameters is (0.0309, 0.203) inch\(^2\).

Key concepts

Chi-Square Test

The chi-square test is a statistical method used to determine if there is a significant difference between the expected and observed data. In the context of variance hypothesis testing, it plays a crucial role in deciding if the variance of a sample differs from a claimed population variance. This is done by comparing the test statistic to a critical value from the chi-square distribution table. The formula for the chi-square test statistic is:
\[\chi^2 = \frac{(n - 1)s^2}{\sigma_0^2}\]
where:

• \(n\) is the sample size,
• \(s^2\) is the sample variance,
• \(\sigma_0^2\) is the population variance under the null hypothesis.

This method allows us to test the hypothesis that there is no difference between the sample variance and the population variance. If the calculated chi-square statistic is greater than the critical value, we reject the null hypothesis, indicating a significant difference.

Confidence Interval

A confidence interval provides a range of values that likely contain the true population parameter. For the variance in our exercise, we calculate a confidence interval to better estimate the true variance of the rod diameters.

The calculation involves finding both the lower and upper bounds from the chi-square distribution. The formulas used are:

• Lower limit: \(\frac{(n-1) s^2}{\chi^2_\text{upper}}\)
• Upper limit: \(\frac{(n-1) s^2}{\chi^2_\text{lower}}\)

Here, \(\chi^2_\text{upper}\) and \(\chi^2_\text{lower}\) are the chi-square values corresponding to the upper and lower tails of the desired confidence level. The resulting interval gives us a range in which the true variance is likely to fall, adding reliability to our statistical findings.

Significance Level

The significance level, often denoted by \(\alpha\), represents the probability of rejecting the null hypothesis when it is true. It is a crucial part of conducting a hypothesis test, as it defines the threshold for determining statistical significance. In this problem, a common significance level of \(\alpha = 0.05\) is used.

Using a significance level of 0.05 implies that there is a 5% risk of incorrectly rejecting the null hypothesis (known as a Type I error). Selecting an appropriate \(\alpha\) level involves balancing the risk of Type I errors with the need to detect meaningful effects or differences when they exist, helping researchers make confident conclusions about their statistical tests.

Degrees of Freedom

Degrees of freedom (df) are essential in determining the critical values from statistical tables such as the chi-square distribution. They are determined by subtracting the number of estimated parameters from the number of observations. In the context of variance testing, the formula is:

\( df = n - 1 \)
where \( n \) represents the sample size.

Degrees of freedom reflect the number of independent values that can vary in an analysis without violating constraints. They influence the shape of the chi-square distribution curve, which is crucial for deciding where the test statistic lies relative to critical boundaries. In hypothesis testing, using the correct degrees of freedom ensures accurate critical values for valid conclusions.