91Ó°ÊÓ

Chapter 7 presented a CI for the variance \(\sigma^{2}\) of a normal population distribution. The key result there was that the rv \(\chi^{2}=(n-1) S^{2} / \sigma^{2}\) has a chi-squared distribution with \(n-1\) df. Consider the null hypothesis \(H_{0}: \sigma^{2}=\sigma_{0}^{2}\) (equivalently, \(\left.\sigma=\sigma_{0}\right)\). Then when \(H_{0}\) is true, the test statistic \(\chi^{2}=(n-1) S^{2} / \sigma_{0}^{2}\) has a chi-squared distribution with \(n-1\) df. If the relevant alternative is \(H_{\mathrm{a}}: \sigma^{2}>\sigma_{0}^{2}\), rejecting \(H_{0}\) if \((n-1) s^{2} / \sigma_{0}^{2} \geq \chi_{\alpha, n-1}^{2}\) gives a test with significance level \(\alpha\). To ensure reasonably uniform characteristics for a particular application, it is desired that the true standard deviation of the softening point of a certain type of petroleum pitch be at most \(.50^{\circ} \mathrm{C}\). The softening points of ten different specimens were determined, yielding a sample standard deviation of \(.58^{\circ} \mathrm{C}\). Does this strongly contradict the uniformity specification? Test the appropriate hypotheses using \(\alpha=.01\).

Step by step solution

01

Define the Null and Alternative Hypothesis

The null hypothesis for our test is that the variance of the distribution is equal to a specified value: \(H_{0}: \sigma^{2} = 0.25\), as it is given that the standard deviation \(\sigma\) should be at most \(0.5^{\circ}C\). The alternative hypothesis, based on the problem, is \(H_{a}: \sigma^{2} > 0.25\). This represents the situation where the variance is greater than what is claimed.

02

Gather Required Information

We are given that the sample size \(n=10\), the sample standard deviation \(s = 0.58^{\circ}C\), and that the significance level \(\alpha = 0.01\). We are also informed that the test statistic follows a chi-squared distribution with \(n-1 = 9\) degrees of freedom.

03

Calculate the Test Statistic

The test statistic for this hypothesis test is calculated as \[ \chi^{2} = \frac{(n-1)s^2}{\sigma_{0}^{2}} = \frac{(10-1)(0.58)^2}{0.25} = \frac{9 \cdot 0.3364}{0.25} = 12.1104. \]

04

Determine the Critical Value

We need the critical value from the chi-squared distribution with 9 degrees of freedom at \(\alpha = 0.01\). Using the chi-squared distribution table or an appropriate statistical tool, we find \(\chi^{2}_{0.01, 9} = 21.666\). This is the critical value.

05

Compare Test Statistic to Critical Value

The calculated \(\chi^{2}\) test statistic is 12.1104, which is less than the critical value of 21.666.

06

Decision and Conclusion

Since the test statistic \(12.1104\) is less than the critical value \(21.666\), we fail to reject the null hypothesis \(H_{0}\). This means there is not enough evidence at the 0.01 significance level to conclude that the standard deviation exceeds 0.50 degrees Celsius.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-Squared Distribution

The chi-squared distribution is a probability distribution commonly used in statistics, especially in hypothesis testing and variance analysis. It's essential when dealing with data that follows a normal distribution. This distribution is asymmetric and only takes positive values. The shape of the chi-squared distribution depends on the degrees of freedom (df), which is determined by the sample size.

• In the context of our hypothesis test, we use the formula \[ \chi^{2} = \frac{(n-1)S^2}{\sigma^2} \] where \(n-1\) degrees of freedom occur because we estimate the population variance from the sample variance.
• The larger the value of \(n-1\), the more the distribution becomes symmetric, resembling a normal distribution.
• The chi-squared distribution is widely used for tests concerning sample variances and to check goodness of fit for models.

Significance Level

The significance level, denoted by \( \alpha \), is a threshold that determines when we will reject the null hypothesis in a hypothesis test. It reflects the probability of committing a Type I error, which is rejecting the true null hypothesis.

• A common choice for \( \alpha \) is 0.05, but in this case, we're using a more stringent level of 0.01.
• The significance level sets the critical value for our test, which is derived from the chi-squared distribution with specific degrees of freedom.
• This critical value helps us decide if our test statistic is extreme enough to reject the null hypothesis.

Selecting a smaller \( \alpha \) (like 0.01) means we're being more conservative, reducing the chances of a Type I error but increasing the chances of a Type II error.

Null and Alternative Hypotheses

The null and alternative hypotheses are foundational in hypothesis testing. They represent two competing claims about a population parameter based on sample statistics.

• The null hypothesis \( H_0 \) presumes there is no effect or difference. In our exercise, this is \( H_0: \sigma^2 = 0.25 \), implying the sample variance is as expected.
• The alternative hypothesis \( H_a \) suggests there is a significant effect or difference. Here, it is \( H_a: \sigma^2 > 0.25 \), indicating the variance is greater than expected.
• The test's goal is to determine whether to reject \( H_0 \) based on the evidence provided by the test statistic.

Hypotheses testing enables researchers to make inferences about population attributes using sample data.

Variance Testing

Variance testing is essential for understanding the variation within a dataset. It helps determine if observed differences in sample data reflect actual differences in populations or if they're due to random variability.

• The test assesses whether the variance of a single sample significantly differs from a hypothesized population variance.
• This utilizes the chi-squared statistic calculated as \[ \chi^{2} = \frac{(n-1)s^2}{\sigma_{0}^{2}} \]where \( s^2 \) is the sample variance and \( \sigma_{0}^{2} \) is the hypothesized population variance.
• A key application of variance testing is quality control, ensuring that products or processes meet acceptable variation levels.

In our example, variance testing helps to decide if the actual variance exceeds the specified threshold, ensuring product consistency.