9. Hypothesis Testing for One Sample

Performing Hypothesis Tests: Variance

9. Hypothesis Testing for One Sample

Performing Hypothesis Tests: Variance: Videos & Practice Problems

Topic summary

Hypothesis testing for population variance uses the chi-square (χ²) statistic calculated by $χ ² = \frac{(n - 1) s ²}{σ}$ , where $n$ is sample size, $s ²$ is sample variance, and $σ ²$ is population variance. The test compares the p-value to the alpha level to decide on rejecting the null hypothesis. Conditions include a random sample and normally distributed data. This method is essential for assessing variance claims in quality control and statistical inference.

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Performing Hypothesis Tests: Variance

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Performing Hypothesis Tests: Variance Video Summary

Testing a claim about a population variance, denoted as σ², involves a hypothesis test that uses the chi-square (χ²) distribution. This process follows the familiar steps of hypothesis testing but requires a specific test statistic formula tailored for variance. For example, consider a scenario where a cereal packaging line requires the fill weight variance to be no greater than 0.25 grams squared. A sample of 30 boxes yields a sample variance of 0.31 grams squared, and the goal is to test if the population variance exceeds 0.25 at a significance level of α = 0.1, assuming the fill weights are normally distributed.

The first step is to establish the hypotheses. The null hypothesis (H₀) states that the population variance equals the specified value: $H_0: \sigma^2 = 0.25$. The alternative hypothesis (H₁) reflects the claim to be tested, which in this case is that the variance is greater than 0.25: $H_1: \sigma^2 > 0.25$.

Next, the test statistic 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 $\sigma_0^2$ is the hypothesized population variance under the null hypothesis. For the example, with $n = 30$, $s^2 = 0.31$, and $\sigma_0^2 = 0.25$, the test statistic is:

\[\chi^2 = \frac{(30 - 1) \times 0.31}{0.25} = \frac{29 \times 0.31}{0.25} \approx 44.59\]

The degrees of freedom for this test are $df = n - 1 = 29$. Since the alternative hypothesis is one-sided (greater than), the p-value corresponds to the right-tail probability of the chi-square distribution with 29 degrees of freedom at the calculated test statistic value. Using chi-square tables or technology, the p-value is approximately 0.032.

Comparing the p-value to the significance level, \$0.032 < 0.1$, leads to rejecting the null hypothesis. This indicates sufficient evidence to support the claim that the population variance exceeds 0.25 grams squared.

It is essential to verify the assumptions underlying this test: the sample must be random, and the data should be normally distributed. In this example, the sample is assumed random, and the problem states the fill weights are normally distributed, satisfying these conditions. Meeting these assumptions ensures the validity of the chi-square test for variance.

Understanding how to perform hypothesis tests for population variance expands statistical inference beyond means and proportions, allowing for comprehensive analysis of variability within populations. Mastery of this method involves recognizing when to apply the chi-square distribution, correctly formulating hypotheses, calculating the test statistic, and interpreting results within the context of the problem.

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 $=0.0016<0.01=\alpha$ , we fail to reject $H_0$ , not enough evidence to suggest $\sigma>0.04$ .

No need to service the machine.

Since $P$ -value $=7.59\times10^{-4}<0.01=\alpha$ , we reject $H_0$ , enough evidence to suggest $\sigma>0.04$ .

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 previous data indicates a population standard deviation of 6.3 percentage points, the population variance ($\sigma^2$) is calculated as \$6.3^2 = 39.69\(. However, the problem states the population variance as 36.69, which we will use as the expected value.

To test if the variance has changed, the null hypothesis (\)H_0$) assumes that the population variance remains at 36.69, while the alternative hypothesis ($H_a\() posits that the variance is different, expressed as:

\[H_0: \sigma^2 = 36.69 \\H_a: \sigma^2 \neq 36.69\]

This two-tailed test uses a significance level of \)\alpha = 0.1$. The sample consists of 40 students, with a sample standard deviation ($s$) of 5.9, leading to a sample variance ($s^2\() of:

\[s^2 = 5.9^2 = 34.81\]

The test statistic for variance follows a chi-square distribution and is calculated by:

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

Substituting the values:

\[\chi^2 = \frac{(40 - 1) \times 34.81}{36.69} = \frac{39 \times 34.81}{36.69} \approx 37.002\]

The degrees of freedom for this test is \)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.

Using chi-square tables or technology, the right-tailed probability for \)\chi^2 = 37.002\( with 39 degrees of freedom is approximately 0.56. The left-tailed probability is then:

\[1 - 0.56 = 0.44\]

The smaller tail probability is 0.44, so the two-tailed p-value is:

\[2 \times 0.44 = 0.88\]

Comparing the p-value to the significance level \)\alpha = 0.1$, since $0.88 > 0.1$, we fail to reject the null hypothesis. This indicates insufficient evidence to conclude that the variance of final exam scores has changed from the historical value of 36.69.

Understanding how to perform a chi-square test for variance is crucial when assessing changes in variability within normally distributed data. This process involves formulating appropriate hypotheses, calculating the test statistic using sample variance and population variance, determining degrees of freedom, and interpreting the p-value in the context of the chosen significance level.

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To perform a hypothesis test on population variance, we use the chi-square test statistic formula: $χ test = \frac{(n - 1) s^{2}}{σ^{2}}$ , 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. We compare the calculated chi-square value to critical values or use the p-value to decide whether to reject the null hypothesis about the population variance.

When performing a hypothesis test for population variance, two main conditions must be met. First, the sample should be random, meaning each observation is independently and randomly selected from the population. This ensures the sample represents the population well. Second, the data must be normally distributed because the chi-square test for variance assumes normality. If the data is not normal, the test results may not be valid. Meeting these conditions allows us to confidently use the chi-square distribution to test claims about the population variance.

In a hypothesis test for 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 (alpha), we 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 alpha, we fail to reject the null hypothesis, meaning there is not enough evidence to conclude the variance differs from the hypothesized value.

The chi-square distribution is used in hypothesis testing for population variance because the test statistic, which involves the sample variance scaled by the hypothesized variance and degrees of freedom, follows a chi-square distribution when the data is normally distributed. Specifically, the formula $χ test = \frac{(n - 1) s^{2}}{σ^{2}}$ follows a chi-square distribution with $n - 1$ degrees of freedom. This property allows us to calculate probabilities and make decisions about the population variance.

When setting up hypotheses for a variance test, the null hypothesis ( $H 0$ ) usually states that the population variance equals a specific value: $σ_{2} = σ 02$ . The alternative hypothesis ( $H a$ ) depends on the claim being tested and can be one-sided or two-sided: $σ_{2} \neq σ 02$ (not equal), $σ_{2} > σ 02$ (greater than), or $σ_{2} < σ 02$ (less than). This setup guides the direction of the test and the calculation of p-values.

In a hypothesis test for population variance, the degrees of freedom (df) are calculated as the sample size minus one: $df = n - 1$ . This is because the sample variance is based on $n$ observations but uses the sample mean, which reduces the independent pieces of information by one. The degrees of freedom are essential for determining the appropriate chi-square distribution to use when finding critical values or p-values.