Question

Independent random samples from two normal populations produced the given variances: $$ \begin{array}{cc} \text { Sample Size } & \text { Sample Variance } \\ \hline 13 & 18.3 \\ 13 & 7.9 \end{array} $$ a. Do the data provide sufficient evidence to indicate that \(\sigma_{1}^{2}>\sigma_{2}^{2}\) ? Test using \(\alpha=.05\). b. Find the approximate \(p\) -value for the test and interpret its value.

Short answer

Answer: No, there is not enough evidence to claim that the variance of the first population is greater than the variance of the second population, as we failed to reject the null hypothesis and the p-value (0.0514) is greater than the level of significance (0.05).

Step-by-step solution

State the null and alternative hypotheses

The null hypothesis states that there is no difference in the variances of the two populations. The alternative hypothesis states that the variance of the first population is greater than the variance of the second population. Mathematically, these hypotheses can be expressed as: - Null hypothesis (H0): \(\sigma_{1}^{2} = \sigma_{2}^{2}\) - Alternative hypothesis (H1): \(\sigma_{1}^{2} > \sigma_{2}^{2}\)

Find the test statistic and critical value

We will use the F-test statistic to compare the sample variances. The formula for the F-test statistic is: $$F = \frac{s_{1}^{2}}{s_{2}^{2}}$$ where \(s_{1}^{2}\) and \(s_{2}^{2}\) are the sample variances. Plugging in the given values, we get: $$F = \frac{18.3}{7.9} \approx 2.316$$ Now, we need to find the critical value for the F-distribution. We are given a level of significance, \(\alpha = 0.05\). Since we're testing whether \(\sigma_{1}^{2} > \sigma_{2}^{2}\), this is a one-tailed test. The degrees of freedom for the first sample (numerator) are \(n_{1} - 1 = 12\), and the degrees of freedom for the second sample (denominator) are \(n_{2} - 1 = 12\). Therefore, the critical value can be found using the F-distribution table or a calculator: $$F_{critical} = F_{0.95, 12, 12} \approx 2.805$$

Make a decision based on the test statistic and critical value

We compare the test statistic to the critical value: - If the test statistic is greater than the critical value, then we reject the null hypothesis and conclude that there is sufficient evidence to support the alternative hypothesis. - If the test statistic is less than or equal to the critical value, then we fail to reject the null hypothesis and cannot conclude that there is sufficient evidence to support the alternative hypothesis. In this case, our test statistic (F = 2.316) is less than the critical value (2.805), so we fail to reject the null hypothesis.

Calculate the p-value and interpret it

Now, we will determine the p-value for the F-test. The p-value is the probability of obtaining a test statistic as extreme or more extreme than the one calculated, given that the null hypothesis is true. To find the p-value, we can use an F-distribution calculator. For our test statistic F = 2.316, and degrees of freedom 12 and 12, we find that the p-value is approximately 0.0514. Since the p-value (0.0514) is greater than the level of significance (\(\alpha = 0.05\)), we fail to reject the null hypothesis. This means we cannot conclude that there is sufficient evidence to indicate that \(\sigma_{1}^{2} > \sigma_{2}^{2}\). The p-value of 0.0514 indicates that there is a 5.14% chance of observing a test statistic as extreme or more extreme than 2.316 when the null hypothesis is true.

Key concepts

Hypothesis Testing

Hypothesis testing is a statistical method that helps us make decisions or draw conclusions about a population based on sample data. It begins by assuming a null hypothesis, which is a statement of no effect or no difference in the context of the specific test. We then compare this assumption against an alternative hypothesis, which represents what we suspect might be true.
In the context of variance comparison using an F-test, the null hypothesis (\(H_0\) : \(\sigma_1^2 = \sigma_2^2\)) states that the population variances are equal. The alternative hypothesis (\(H_1\) : \(\sigma_1^2 > \sigma_2^2\)) suggests that the variance of the first population is greater than that of the second.

• The null hypothesis always contains an equality.
• The alternative hypothesis indicates the presence of an effect or difference.

The decision to reject or not reject the null hypothesis is based on statistical evidence collected from the sample data. This decision is supported by calculating a test statistic and comparing it to a critical value from a statistical distribution or by determining a p-value.
A rejection of the null hypothesis suggests there is enough evidence to support the alternative hypothesis, indicating a significant effect or difference in the population being tested.

P-value

The p-value is a key concept in hypothesis testing, representing the probability of obtaining a test statistic as extreme as the observed one, assuming that the null hypothesis is true. It essentially tells us how extreme the sample data is. In the context of the F-test for variance comparison:

• If the p-value is less than the alpha level (\(\alpha = 0.05\)), we reject the null hypothesis, implying there is sufficient evidence to conclude that the population variances are different.
• If the p-value is greater than \(\alpha\), we fail to reject the null hypothesis, indicating insufficient evidence to support the alternative claim.

In our example, the calculated p-value (approximately 0.0514) is slightly greater than the alpha level of 0.05. This means our result is not statistically significant at the 5% level, and we cannot conclude that the variance of the first population is greater than that of the second. The p-value provides a measure of the evidence against the null hypothesis; a lower p-value represents stronger evidence in favor of the alternative hypothesis.

Variance Comparison

Variance comparison is an essential aspect of statistical analysis, especially when trying to determine if two or more groups differ in variability. In many practical situations, comparing variances helps understand how different groups respond to various conditions or treatments.
The F-test is commonly used for this purpose. It evaluates two sample variances to infer about the population variances.

• The formula for the F-test statistic is \(F = \frac{s_1^2}{s_2^2}\), where \(s_1^2\) and \(s_2^2\) are the sample variances.
• The calculated F value is then compared to a critical value from the F-distribution.

Evaluating the variances of two independent samples involves understanding if their spread or dispersion significantly differs. In this exercise, the F value of 2.316 came from dividing the higher sample variance (18.3) by the lower one (7.9), to test if the variability in the first group is indeed larger.
The decision-making process hinges on whether the calculated F value exceeds the critical value from the F-distribution table, based on the specified significance level (such as \(\alpha = 0.05\)). If the F statistic is smaller than the critical value, we fail to reject the null hypothesis, indicating no significant difference in variances between the two groups.