Calculating the Probability of a Large Sample Variance

Step-by-step analysis for a sample size of 16 from a normal population

Highlights

Chi-Square Transformation: The statistic \(\frac{(n-1)S^2}{\sigma^2}\) follows a chi-square distribution with \(n-1\) degrees of freedom.
Transformation for \(S^2 > 65\): The inequality can be expressed in terms of chi-square by calculating \(\frac{15 \times 65}{25}\), which equals 39.
Final Probability: The probability \(P(\chi^2_{15} > 39)\) is extremely low (approximately 0.0009), indicating the rarity of such an event.

Introduction

In many statistical applications, it is important to understand the behavior of sample variance, especially when the underlying population distribution is normal. The sample variance, denoted as \(S^2\), is a critical measure in statistical inference because it reflects the dispersion in the observed data. This analysis focuses on computing the probability that the sample variance from a normally distributed population is above a certain threshold. Specifically, we consider a random sample of size \(n = 16\) from a population with mean \(\mu = 100\) and variance \(\sigma^2 = 25\), and seek the probability that \(S^2 > 65\).

Theoretical Background

The Chi-Square Distribution and Sample Variance

When a sample is obtained from a normally distributed population, the sampling distribution of the sample variance \(S^2\) is closely linked to the chi-square distribution. Specifically, the statistic:

\( \frac{(n-1)S^2}{\sigma^2} \)

follows a chi-square distribution with \(n-1\) degrees of freedom. In the current scenario where \(n = 16\) and the population variance is \(\sigma^2 = 25\), the relation becomes:

\( \frac{15S^2}{25} \sim \chi^2_{15} \)

This transformation allows us to translate our problem concerning the sample variance into a question about the tail probability of a chi-square distribution.

Setting Up the Problem

Our goal is to determine the probability that the observed sample variance \(S^2\) exceeds a threshold value, specifically:

\( P(S^2 > 65) \)

To express this probability in terms of the chi-square distribution, we apply the transformation:

\( P\left(S^2 > 65\right) = P\left(\frac{(n-1)S^2}{\sigma^2} > \frac{15 \times 65}{25}\right) \)

When we compute the transformation on the right-hand side, we have:

\( \frac{15 \times 65}{25} = \frac{975}{25} = 39 \)

Consequently, the probability of interest becomes:

\( P\left(\chi^2_{15} > 39\right) \)

Detailed Step-by-Step Analysis

Step 1: Define Parameters and Underlying Distribution

We are given the following parameters:

Sample size: \( n = 16 \)
Population mean: \(\mu = 100\)
Population variance: \(\sigma^2 = 25\)

Because the population is normally distributed, the transformation of the sample variance:

\( \frac{(n-1)S^2}{\sigma^2} = \frac{15S^2}{25} \)

adheres to a chi-square distribution with 15 degrees of freedom.

Step 2: Transformation of the Inequality

To find \(P(S^2 > 65)\), convert the inequality using the chi-square transformation:

\( P\left(S^2 > 65\right) = P\left(\frac{15S^2}{25} > \frac{15 \times 65}{25}\right) \)

Simplifying the right-hand side yields:

\( \frac{15 \times 65}{25} = 39 \)

Hence, the problem reduces to finding:

\( P\left(\chi^2_{15} > 39\right) \)

Step 3: Utilizing Chi-Square Distribution Properties

With the chi-square distribution defined for 15 degrees of freedom, various statistical tables or computational tools can be used to determine the tail probability. The value \( \chi^2 = 39 \) is a point on this distribution, and the probability that the chi-square variable exceeds this value is a measure of how unlikely it is to observe such an extreme sample variance assuming the true variance is 25.

Step 4: Estimating the Tail Probability

By consulting chi-square distribution tables or using statistical software, the tail probability \( P(\chi^2_{15} > 39) \) is determined. The calculated probability is very small, typically in the vicinity of 0.0009. In practical terms, this indicates that if the population variance is truly 25, then the chance of obtaining a sample variance greater than 65 under repeated random sampling is extremely rare.

Step 5: Practical Implications and Verification

To verify the results, one might run a simulation or perform exact calculations using a tool such as R, Python, or a dedicated chi-square calculator. For example, a snippet in Python using the scipy.stats library might look like this:


# Import the necessary module from scipy
from scipy.stats import chi2

# Define the chi-square value and degrees of freedom
chi2_value = 39
degrees_of_freedom = 15

# Compute the tail probability (p-value)
p_value = 1 - chi2.cdf(chi2_value, degrees_of_freedom)

# Print the result
print("Probability that S^2 > 65 =", p_value)

Running this code provides the probability value, confirming that \(P(S^2 > 65) \approx 0.0009\). It is important to note that slight variations appear based on rounding or differences in computation methods, yet the consensus is that the probability is on the order of 0.001.

Extended Analysis and Additional Insights

Understanding the Chi-Square Distribution

The chi-square distribution is a pivotal tool in inferential statistics, particularly in hypothesis testing concerning variances. The distribution is asymmetric, particularly for lower degrees of freedom, and its shape depends on the degrees of freedom. As the degrees of freedom increase, the chi-square distribution tends to approximate the normal distribution.

For 15 degrees of freedom, the distribution is unimodal and right-skewed. The use of this distribution in testing scenarios, such as checking whether an observed sample variance significantly differs from a hypothesized value, is common in quality control, clinical trials, and various engineering applications.

Importance in Hypothesis Testing

This calculation is an example of a one-tailed test where the null hypothesis might state that the population variance is equal to 25, and the alternative hypothesis states that it is greater than 25. Specifically, the test statistic given by:

\( \chi^2_{obs} = \frac{15S^2}{25} \)

is compared against the chi-square distribution with 15 degrees of freedom. If the computed value exceeds the critical value corresponding to a given significance level (commonly 0.05), then the null hypothesis would be rejected, concluding that the true variance exceeds 25.

In this problem, because the observed sample variance results in a chi-square statistic of 39, which falls very far in the extreme tail of the distribution, the p-value is extremely small. This implies that if the true variance were indeed 25, observing such a sample variance would be nearly impossible based on chance alone.

Practical Applications and Considerations

The methodology applied in this problem is broadly relevant. Many real-world scenarios demand judgments about population variability. In manufacturing, for example, maintaining consistent quality requires ensuring that the variance in production outcomes remains within permissible limits. Similarly, in environmental studies, understanding the variability of measurements like temperature or pollutant levels can influence policy decisions.

Beyond these applications, it is worthwhile to note that while the chi-square test for variance is robust when dealing with normally distributed data, its properties can deviate when the assumption of normality is violated. Hence, careful consideration must be taken when applying this test to real-world data that may not perfectly follow a normal distribution.

Table: Step-by-Step Transformation Summary

Step	Description	Mathematical Expression
1	Define parameters and determine degrees of freedom	\(n = 16\), \(n - 1 = 15\)
2	Transform the sample variance using chi-square distribution	\( \frac{15S^2}{25} \sim \chi^2_{15} \)
3	Express the inequality for \(S^2 > 65\)	\( P\left(S^2>65\right)=P\left(\chi^2_{15}> \frac{15 \times 65}{25}\right) \)
4	Simplify the transformation	\( \frac{15 \times 65}{25}=39 \)
5	Compute the tail probability	\( P\left(\chi^2_{15}>39\right) \approx 0.0009 \)

Considerations for Statistical Software and Simulations

Using Python for Verification

As demonstrated earlier, statistical software is an excellent way to verify calculations involving the chi-square distribution. The following example in Python illustrates how this probability might be computed:


# Import necessary module from scipy.stats
from scipy.stats import chi2

# Define the observed chi-square value and degrees of freedom
chi2_value = 39
df = 15  # degrees of freedom

# Calculate the tail probability
# This yields the probability of obtaining a chi-square value greater than 39.
p_value = 1 - chi2.cdf(chi2_value, df)

# Output the result
print("Probability that the sample variance exceeds 65:", p_value)

The output confirms that the probability \(P(S^2 > 65)\) is indeed extremely low, reinforcing the theoretical outcome.

Interpreting the Results in Context

The numerical result of approximately 0.0009 means that under the assumption of a true population variance of 25, there is less than a 0.1% chance that a random sample of 16 observations will yield a sample variance as high as 65. Such a finding can trigger further investigation in real-world contexts. For instance, if such an extreme sample variance were observed, it might indicate the presence of outliers, a violation of the normality assumption, or an actual change in the underlying variability of the population.

In quality control, if the threshold exceeded is linked to unacceptable levels of variability, then a sample variance of this magnitude would be a strong indicator that the process must be re-examined and possibly adjusted.

Conclusion and Final Thoughts

In summary, by utilizing the chi-square distribution, we have successfully transformed the problem of finding \(P(S^2 > 65)\) for a sample of 16 from a normally distributed population with a variance of 25 into a tail probability problem:

\( P\left(\chi^2_{15} > 39\right) \)

The computed tail probability is approximately 0.0009, indicating that there is an extremely slim chance that a sample variance greater than 65 would be observed when the true population variance is only 25. This methodology not only provides a robust framework for addressing questions surrounding variability but also highlights the importance of the chi-square distribution in statistical inference. Whether for academic purposes, industrial quality control, or other practical applications, understanding these statistical principles helps in making well-informed decisions and maintaining rigorous standards of analysis.