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, \(\sigma=\sigma_{0}\). Then when \(H_{0}\) is true, the test statistic \(\chi^{2}=(n-1) \quad S^{2} / \sigma_{0}^{2}\) 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

Identify Hypotheses

For this problem, the null hypothesis is \( H_0: \sigma^2 = \sigma_0^2 \), which means \( \sigma = 0.50 \text{ °C} \). The alternative hypothesis is \( H_a: \sigma^2 > \sigma_0^2 \), indicating that the standard deviation is greater than \( 0.50 \text{ °C} \).

02

Gather Data and Parameters

We have a sample size \( n = 10 \) and a sample standard deviation of \( s = 0.58 \text{ °C} \). The level of significance \( \alpha = 0.01 \).

03

Calculate Test Statistic

Use the formula \( \chi^2 = \frac{(n-1)s^2}{\sigma_0^2} \) to compute the test statistic. Here, \( \sigma_0 = 0.50 \text{ °C} \), so the test statistic is \[ \chi^2 = \frac{(10-1) \times (0.58)^2}{(0.50)^2} = \frac{9 \times 0.3364}{0.25} = \frac{3.0276}{0.25} = 12.1104 \].

04

Find Critical Value

Refer to the chi-square distribution table with \( df = n - 1 = 9 \) for \( \alpha = 0.01 \). The critical value \( \chi^2_{\alpha,n-1} \) is approximately 21.67.

05

Compare Test Statistic with Critical Value

The calculated test statistic \( \chi^2 = 12.1104 \) is less than the critical value 21.67. Therefore, we do not reject the null hypothesis \( H_0 \).

06

Conclude the Hypothesis Test

Since the test statistic does not exceed the critical value, we do not have enough evidence to say that the true standard deviation of the softening points is greater than \( 0.50 \text{ °C} \) at a \( 0.01 \) significance level.

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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 vital concept in hypothesis testing, especially when dealing with variance or standard deviation. It is a distribution that emerges when you square standard normal variables and sum them up. Chi-squared distribution is denoted as \(\chi^2\) and is used extensively because of its ability to model the distribution of sample variance.

This distribution is characterized by its degrees of freedom (df). The degrees of freedom typically correspond to the sample size minus one, represented as \(n-1\). In hypothesis testing, it helps us understand how likely or unlikely a test statistic is under the null hypothesis.

• For our problem, the chi-squared statistic is computed using the formula \(\chi^2 = \frac{(n-1)S^2}{\sigma_0^2}\).
• The test statistic follows a chi-squared distribution with \(n-1\) degrees of freedom.

The chi-squared distribution provides the critical value, against which the test statistic is compared to decide whether to reject or not reject the null hypothesis.

Null Hypothesis

The null hypothesis, commonly denoted as \(H_0\), is a foundational idea in statistical hypothesis testing. It functions as a starting assumption or statement that there is no effect or no difference, which the hypothesis test seeks to support or reject.

In the problem context, the null hypothesis states that the variance \(\sigma^2\) of the petroleum pitch's softening point is equal to a specific variance \(\sigma_0^2\), which translates to a standard deviation of 0.50 °C.

• Symbolically, this is expressed as \(H_0: \sigma^2 = \sigma_0^2\).
• The null hypothesis posits that the observed data arises from a population with this variance.

Testing this null hypothesis with sample data involves calculating a test statistic and comparing it to a critical value derived from a chi-squared distribution. If the statistic exceeds the critical value, the null hypothesis is rejected, indicating a statistically significant difference.

Significance Level

The significance level, denoted as \(\alpha\), is a threshold used in hypothesis testing that determines when to reject the null hypothesis. It represents the probability of making a Type I error - rejecting the null hypothesis when it is actually true.

In this scenario, the significance level is set at \(0.01\), meaning there is a 1% risk of concluding that the pitch’s standard deviation is greater than 0.50 °C when in fact, it is not.

• A lower \(\alpha\) value indicates a stricter criterion for rejecting \(H_0\).
• Critical values are derived from the chi-squared distribution table for the chosen \(\alpha\) and within the context of the degrees of freedom.

For this hypothesis test, the significance level helps define the critical region. If the test statistic calculated from the sample data falls within this critical region, the null hypothesis is rejected.

Alternative Hypothesis

The alternative hypothesis, denoted \(H_a\), offers a different perspective to the null hypothesis. It represents the outcome that the test is attempting to provide evidence for, essentially stating there is an effect or difference.

In this exercise, the alternative hypothesis is that the variance \(\sigma^2\) of the petroleum pitch's softening point is greater than the specified variance \(\sigma_0^2\), leading to a standard deviation exceeding 0.50 °C.

• The alternative hypothesis is symbolized as \(H_a: \sigma^2 > \sigma_0^2\).
• This hypothesis is only considered if the null hypothesis is rejected.

The aim of the hypothesis test is to establish enough evidence to reject the null hypothesis in favor of the alternative, suggesting that the true standard deviation indeed exceeds the desired specification.