What are the steps involved in performing a hypothesis test for population variance?

Performing a hypothesis test for population variance involves four main steps. First, state the null hypothesis H 0 and alternative hypothesis H a about the population variance σ 2 . Second, calculate the test statistic using the chi-square formula χ 2 = ( n - 1 ) s 2 σ 2 . Third, determine the p-value using the chi-square distribution with n - 1 degrees of freedom. Finally, compare the p-value to the significance level α to decide whether to reject or fail to reject the null hypothesis.

9. Hypothesis Testing for One Sample

Performing Hypothesis Tests: Variance

9. Hypothesis Testing for One Sample

Performing Hypothesis Tests: Variance: Videos & Practice Problems

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Hypothesis testing for population variance involves testing claims about $σ^{2}$ using a chi-square test statistic calculated by $χ calc = \frac{(n - 1) s^{2}}{σ^{2}}$ . The degrees of freedom are $n - 1$ . Comparing the p-value to the significance level $α$ determines whether to reject the null hypothesis $H 0$ . Assumptions include a random sample and normally distributed data, ensuring valid inferential statistics for variance analysis.

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Performing Hypothesis Tests: Variance

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Performing Hypothesis Tests: Variance Video Summary

Testing hypotheses about a population variance involves a process similar to tests for means or proportions but uses a chi-square test statistic. This method is essential when you want to determine if the variability in a population differs from a specified value, such as in quality control scenarios where variance limits are critical.

To conduct a hypothesis test for population variance, start by defining the null hypothesis (H₀) as the population variance (σ²) being equal to a specified value, and the alternative hypothesis (H_a) as the variance being greater than, less than, or not equal to that value, depending on the claim. For example, if a cereal packaging line requires the fill weight variance to be no greater than 0.25 grams squared, and a sample variance is observed, the hypotheses might be:

$H_0: \sigma^2 = 0.25$

$H_a: \sigma^2 > 0.25$

The test statistic for this hypothesis test is calculated using the formula:

\[\chi^2 = \frac{(n - 1) s^2}{\sigma_0^2}\]

where n is the sample size, s² is the sample variance, and σ₀² is the hypothesized population variance under the null hypothesis. This formula leverages the chi-square distribution, which is appropriate because the sampling distribution of the variance of a normally distributed population follows a chi-square distribution with n - 1 degrees of freedom.

After calculating the chi-square test statistic, determine the degrees of freedom as df = n - 1. Then, find the p-value corresponding to the test statistic and the degrees of freedom. The direction of the test (right-tailed, left-tailed, or two-tailed) depends on the alternative hypothesis. For a greater-than alternative, the p-value is the right-tail probability of the chi-square distribution.

Compare the p-value to the significance level (α). If the p-value is less than α, reject the null hypothesis, indicating sufficient evidence that the population variance differs as specified by the alternative hypothesis. For instance, if α = 0.1 and the p-value is approximately 0.032, the null hypothesis is rejected, supporting the claim that the variance exceeds 0.25 grams squared.

It is crucial to verify the assumptions before drawing conclusions. The sample should be randomly selected, and the population data must be approximately normally distributed, as the chi-square test for variance relies on normality. When these conditions are met, the test results are valid and can inform decisions about population variability.

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Problem

A machine produces ball bearings that are designed to have a diameter standard deviation of 0.04 mm, but an engineer suspects the variability has increased. A sample of 60 bearings shows a standard deviation of 0.052 mm. Perform a hypothesis test with $α = 0.01$ to test the claim. Should the line manager have the machine serviced?

Since $P$ -value $=7.59\times10^{-4}<0.01=\alpha$ , we fail to reject $H_0$ , not enough evidence to suggest $\sigma>0.04$ .

No need to service the machine.

Since $P$ -value $= 0.0016 < 0.01 = α$ , we fail to reject $H_0$ , not enough evidence to suggest $\sigma>0.04$ .

No need to service the machine.

Since $P$ -value $= 7.59 \times 10^{- 4} < 0.01 = α$ , we reject $H_{0}$ , enough evidence to suggest $σ > 0.04$ .

Machine should be serviced.

Since $P$ -value $=0.0016<0.01=\alpha$ , we reject $H_0$ , enough evidence to suggest $\sigma>0.04$ .

Machine should be serviced.

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Performing Hypothesis Tests: Variance Example 1

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Performing Hypothesis Tests: Variance Example 1 Video Summary

When analyzing whether the variance of final exam scores has changed after modifying instructional practices, it is essential to conduct a hypothesis test for the population variance. Given that the historical population variance ($\sigma^2$) is 36.69, derived from a standard deviation of 6.3 percentage points, the goal is to determine if the current semester's variance differs significantly. A random sample of 40 students yielded a sample standard deviation ($s$) of 5.9, which corresponds to a sample variance of $s^2 = 5.9^2 = 34.81$.

The hypotheses for this two-tailed test are formulated as follows: the null hypothesis ($H_0$) assumes the population variance remains unchanged, so $\sigma^2 = 36.69$, while the alternative hypothesis ($H_a$) posits that the variance has changed, meaning $\sigma^2 \neq 36.69$. Since the exam scores are normally distributed, the chi-square ($\chi^2$) test statistic is appropriate for testing variance changes. The test statistic is calculated using the formula:

\[\chi^2 = \frac{(n - 1) s^2}{\sigma^2}\]

where $n$ is the sample size. Substituting the values:

\[\chi^2 = \frac{(40 - 1) \times 34.81}{36.69} \approx 37.002\]

The degrees of freedom for this test are $n - 1 = 39$. Since the alternative hypothesis is two-sided, the p-value is determined by doubling the smaller tail probability of the chi-square distribution with 39 degrees of freedom. Using chi-square tables or technology, the right-tailed probability for $\chi^2 = 37.002$ is approximately 0.56, so the left-tailed probability is \$1 - 0.56 = 0.44\(. The smaller tail probability is 0.44, thus the two-tailed p-value is:

\[p = 2 \times 0.44 = 0.88\]

Comparing this p-value to the significance level \)\alpha = 0.1$, since $0.88 > 0.1$, there is insufficient evidence to reject the null hypothesis. This means the data do not support a significant change in the variance of final exam scores. Therefore, the variance remains consistent with previous years, indicating that the instructional changes did not significantly affect score variability.

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9. Hypothesis Testing for One Sample

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When performing a hypothesis test for population variance, the test statistic is calculated using the chi-square distribution. The formula is $χ 2 = \frac{(n - 1) s^{2}}{σ^{2}}$ , where $n$ is the sample size, $s^{2}$ is the sample variance, and $σ^{2}$ is the hypothesized population variance. This test statistic follows a chi-square distribution with $n - 1$ degrees of freedom. This formula helps determine if the observed sample variance significantly differs from the hypothesized variance.

To validly perform a hypothesis test on population variance, two key assumptions must be met. First, the sample must be randomly selected to ensure that it represents the population without bias. Second, the data should be normally distributed because the chi-square test for variance relies on this assumption for accurate inference. If the data are not normal, the test results may not be reliable. Checking these assumptions is crucial before interpreting the test results, as violating them can lead to incorrect conclusions about the population variance.

In a hypothesis test for population variance, the p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from the sample data, assuming the null hypothesis is true. If the p-value is less than the chosen significance level $α$ , you reject the null hypothesis, indicating there is sufficient evidence to support the alternative hypothesis about the population variance. Conversely, if the p-value is greater than $α$ , you fail to reject the null hypothesis, meaning there is not enough evidence to conclude a difference in variance.

Performing a hypothesis test for population variance involves four main steps. First, state the null hypothesis $H 0$ and alternative hypothesis $H a$ about the population variance $σ^{2}$ . Second, calculate the test statistic using the chi-square formula $χ 2 = \frac{(n - 1) s^{2}}{σ^{2}}$ . Third, determine the p-value using the chi-square distribution with $n - 1$ degrees of freedom. Finally, compare the p-value to the significance level $α$ to decide whether to reject or fail to reject the null hypothesis.

The degrees of freedom (df) for a chi-square test on population variance is calculated as the sample size minus one. Mathematically, it is $df = n - 1$ , where $n$ is the number of observations in the sample. This value is used to reference the appropriate chi-square distribution when determining the p-value for the test statistic. Correctly identifying the degrees of freedom is essential for accurate hypothesis testing of variance.

Performing a hypothesis test for population variance assumes that the data are normally distributed. This assumption is critical because the chi-square test statistic relies on normality to follow the chi-square distribution accurately. If the data are not normally distributed, the test results may be invalid or misleading. In such cases, alternative methods or nonparametric tests should be considered. Therefore, before conducting the test, it is important to assess the normality of the data to ensure the validity of the hypothesis test for variance.

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Performing Hypothesis Tests: Variance: Videos & Practice Problems

Performing Hypothesis Tests: Variance

Performing Hypothesis Tests: Variance Video Summary

Performing Hypothesis Tests: Variance Example 1

Performing Hypothesis Tests: Variance Example 1 Video Summary

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What is the formula for the test statistic when performing a hypothesis test for population variance?

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What are the steps involved in performing a hypothesis test for population variance?

How do you find the degrees of freedom for a chi-square test on population variance?

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