Question

Independent random samples from two normal populations produced the variances listed here: $$ \begin{array}{cc} \text { Sample Size } & \text { Sample Variance } \\ \hline 16 & 55.7 \\ 20 & 31.4 \end{array} $$ a. Do the data provide sufficient evidence to indicate that \(\sigma_{1}^{2}\) differs from \(\sigma_{2}^{2}\) ? Test using \(\alpha=.05\). b. Find the approximate \(p\) -value for the test and interpret its value.

Short answer

Explain your answer by mentioning the conclusion from the hypothesis test and the comparison of p-value and significance level.

Step-by-step solution

State the hypotheses

We will write the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)) for this exercise. The null hypothesis would be that the population variances are equal, while the alternative hypothesis would be they are different. Null hypothesis (\(H_0\)): \(\sigma_1^2 = \sigma_2^2\) Alternative hypothesis (\(H_a\)): \(\sigma_1^2 \neq \sigma_2^2\)

Calculate the test statistic

We will use the F-test statistic for testing the hypothesis. 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. Using the given data, \(S_1^2 = 55.7\) and \(S_2^2 = 31.4\) F = \(\frac{55.7}{31.4}\) F ≈ 1.773

Find the critical values and p-value

For this step, we need to find the critical values using the F distribution table based on the given significance level (\(\alpha = 0.05\)) and degrees of freedom. The degrees of freedom are calculated as: df1 = n1 - 1 (for sample 1) df2 = n2 - 1 (for sample 2) Using the given sample sizes, df1 = 16 - 1 = 15 df2 = 20 - 1 = 19 With \(\alpha = 0.05\), and df1 = 15 and df2 = 19, we find the critical values from the F-distribution table: Lower critical value (F1) ≈ 0.470 Upper critical value (F2) ≈ 2.152 We also need to calculate the p-value. Using an F-distribution calculator or relevant software, input the F statistic (F = 1.773) and the degrees of freedom (df1 = 15, df2 = 19). We obtain the p-value: p-value ≈ 0.073

Draw a conclusion based on the p-value and the significance level

Now, we will compare the p-value with the given significance level \(\alpha=0.05\) to decide whether to accept or reject the null hypothesis. Since the p-value (≈ 0.073) is greater than the significance level (0.05), we do not have sufficient evidence to reject the null hypothesis. Therefore, there is insufficient evidence to conclude that the population variances differ. Moreover, the test statistic, F ≈ 1.773, falls between the lower critical value (0.470) and the upper critical value (2.152), which also indicates that we fail to reject the null hypothesis.

Key concepts

F-test

The F-test is a statistical method used to compare two variances to see if they are significantly different from one another. It is called an F-test because it uses the F-distribution, a type of probability distribution that arises when two variances are compared. The test statistic for an F-test is calculated using the formula: \[ F = \frac{S_1^2}{S_2^2} \]Where \(S_1^2\) and \(S_2^2\) are the sample variances. If \(F\) is significantly high or low compared to a critical value from the F-distribution table, it suggests the variances are different.The F-test is used often in Analysis of Variance (ANOVA) and is vital for assessing group variability in stats. This test requires certain assumptions:

• Both samples should come from normal distributions.
• The samples should be independent of each other.
• The ratio of sample variances should be larger than 1.

Understanding F-tests helps ensure the decisions about variance are statistically sound and not by random chance.

Null Hypothesis

The null hypothesis, denoted as \(H_0\), is a statement in hypothesis testing that signifies no effect or no difference. It's often the hypothesis that the experiment seeks to nullify. For example, in our exercise, the null hypothesis is that the population variances are equal: \[H_0: \sigma_1^2 = \sigma_2^2 \]The null hypothesis acts as a starting point for statistical analysis. The aim is usually to collect enough evidence to reject this hypothesis. If the null is rejected, it suggests that the alternative hypothesis \(H_a\) is valid, which in this case would mean that the variances are indeed different. In hypothesis testing, formulating the null hypothesis precisely and understanding its role is crucial. This ensures that statistical interpretations remain clear, consistent, and defensible.

Significance Level

The significance level, denoted by \(\alpha\), is a threshold set by the researcher to determine whether to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true, also known as the Type I error rate. Common choices for the significance level are \(\alpha = 0.05\), \(\alpha = 0.01\), and \(\alpha = 0.10\). For our problem, the significance level is \(0.05\), which suggests a 5% risk of incorrectly rejecting the null hypothesis.Choosing the appropriate \(\alpha\) is essential as it influences the reliability of the test results. A lower significance level requires more substantial evidence against the null hypothesis before it can be rejected. This plays a pivotal role in hypothesis testing by serving as a basis for decision-making.

p-value

The p-value is a crucial element in hypothesis testing that helps determine the significance of the results. It represents the probability of observing a test statistic at least as extreme as the one calculated, assuming the null hypothesis is true. A lower p-value suggests stronger evidence against the null hypothesis.In our exercise, we calculated a p-value of approximately \(0.073\). Since this value is greater than the significance level \(\alpha\) of \(0.05\), we cannot reject the null hypothesis. This means that the observed difference in variances could easily occur by random chance.To interpret the p-value properly:

• If \(p \leq \alpha\): Reject the null hypothesis. Sufficient evidence to support \(H_a\).
• If \(p > \alpha\): Fail to reject \(H_0\). Insufficient evidence to support \(H_a\).

The p-value gives a quantitative measure to assist in decision-making. It's vital for drawing conclusions about the hypothesis' validity.