Confidence Intervals for Population Variance: Videos & Practice Problems

Confidence intervals for variance use the chi-square distribution with critical values $\chi {\sigma }_r$ and $\chi {\sigma }_l$ to calculate bounds: $\mathrm{Lower}\mathrm{bound}=\frac{(n-1)s{²}^{}}{\chi }$ and $\mathrm{Upper}\mathrm{bound}=\frac{(n-1)s{²}^{}}{\chi }$. For standard deviation, take square roots of these bounds. This method assumes normal distributions and uses sample variance as the point estimate, integrating concepts like degrees of freedom and alpha level to find critical chi-square values, enabling precise estimation of population variance and standard deviation within a specified confidence level.

Constructing Confidence Intervals for Variance or Standard Deviation

Constructing a confidence interval for a population variance involves using the sample variance as the point estimate and leveraging the chi-square distribution to determine the interval bounds. Unlike confidence intervals for means, which use a symmetric margin of error based on the t-distribution, the chi-square distribution is asymmetric, so the interval is formed using two critical values from the chi-square table rather than a single margin of error.

To build a confidence interval for variance, start with the sample size n and calculate the degrees of freedom as df = n - 1. The confidence level, denoted as C, helps determine the significance level α = 1 - C. For a two-tailed interval, split this into α/2 for each tail. The critical values, χ²_l and χ²_r, correspond to the chi-square values at the upper and lower tails, found using the degrees of freedom and the probabilities 1 - α/2 and α/2, respectively.

The confidence interval for the population variance σ² is calculated using the formula:

\[\left( \frac{(n - 1) s^2}{\chi^2_r}, \frac{(n - 1) s^2}{\chi^2_l} \right)\]

where s² is the sample variance. Note that the larger chi-square critical value is placed in the denominator of the lower bound, resulting in a smaller lower bound, while the smaller critical value is in the denominator of the upper bound, producing a larger upper bound. This reflects the asymmetry of the chi-square distribution.

For example, if a sample of 12 eggs has a sample variance of 1.2 and a 90% confidence level is desired, the degrees of freedom are 11. The significance level is 0.10, so α/2 = 0.05. Using a chi-square table, the critical values might be approximately χ²_r = 19.68 and χ²_l = 4.58. Plugging these into the formula gives a confidence interval for the variance between approximately 0.67 and 2.89. This means there is 90% confidence that the true variance of egg weights lies within this range.

To find a confidence interval for the population standard deviation σ, simply take the square root of both bounds of the variance interval:

\[\left( \sqrt{\frac{(n - 1) s^2}{\chi^2_r}}, \sqrt{\frac{(n - 1) s^2}{\chi^2_l}} \right)\]

This yields the interval estimate for the standard deviation, maintaining the same confidence level.

It is important to ensure that the underlying population is approximately normally distributed when applying this method, as the chi-square distribution assumptions rely on normality. This approach provides a robust way to estimate the variability of a population based on sample data, using the chi-square distribution to account for the asymmetry inherent in variance estimation.

What is wrong with expressing the confidence interval as ${\sigma }^2=5.1\pm 1.3$?

• A. 5.1 is not the midpoint between 3.8 and 6.4.
• B. The values 3.8 and 6.4 are impossible because variance must be less than 3.
• C. The point estimate for σ² is not the midpoint of a confidence interval and 1.3 is not a margin of error since the $\chi^2$ distribution is asymmetric.
• D. Confidence intervals can only be written for means or proportion, not for variance.

Answer: (C)

A delivery service tracks the weights of its packages. A sample of 20 packages has a variance of 4.5 lbs². Construct a 95% conf. int. for the population variance. Assume a normal distribution.

• A. (2.60, 9.60)
• B. (2.74, 9.60)
• C. (2.99, 8.89)
• D. (2.84, 8.45)

Answer: (A)

Constructing Confidence Intervals for Variance or Standard Deviation Example 1

When estimating the population variance (σ²) from a sample, constructing a confidence interval requires understanding the chi-square distribution, especially when the sample is drawn from a normally distributed population. For a sample size of n = 30 with a sample variance of 1.2 mm², the degrees of freedom (df) is calculated as n - 1 = 29. The confidence interval for the population variance is derived using the chi-square critical values corresponding to the desired confidence level.

To build a confidence interval for variance, the formula used is:

\[\frac{(n - 1) s^2}{\chi^2_{\alpha/2, \, df}} < \sigma^2 < \frac{(n - 1) s^2}{\chi^2_{1 - \alpha/2, \, df}}\]

Here, s² is the sample variance, n - 1 is the degrees of freedom, and χ² values are the chi-square critical values at the specified tail probabilities.

For a 90% confidence interval, the significance level is α = 0.10, so α/2 = 0.05. Using the chi-square distribution table for 29 degrees of freedom, the critical values are approximately χ²_{0.05, 29} = 42.56 and χ²_{0.95, 29} = 17.71. Plugging these into the formula yields:

\[\frac{29 \times 1.2}{42.56} < \sigma^2 < \frac{29 \times 1.2}{17.71}\]\[0.82 < \sigma^2 < 1.96\]

This means we are 90% confident that the true population variance lies between 0.82 and 1.96 mm².

Increasing the confidence level to 99% changes the significance level to α = 0.01, so α/2 = 0.005. The corresponding chi-square critical values for 29 degrees of freedom are approximately χ²_{0.005, 29} = 52.34 and χ²_{0.995, 29} = 13.12. Applying these values:

\[\frac{29 \times 1.2}{52.34} < \sigma^2 < \frac{29 \times 1.2}{13.12}\]\[0.66 < \sigma^2 < 2.65\]

Comparing the two intervals reveals that increasing the confidence level from 90% to 99% results in a wider confidence interval. This expansion reflects the trade-off between confidence and precision: a higher confidence level demands a broader range to ensure the true population variance is captured. The lower bound decreases from 0.82 to 0.66, and the upper bound increases from 1.96 to 2.65, illustrating this effect clearly.

Understanding how confidence levels influence interval width is crucial in quality control and statistical inference, as it balances the certainty of capturing the true parameter against the precision of the estimate.

Constructing Confidence Intervals for Variance or Standard Deviation Example 2

When constructing a confidence interval for the population standard deviation, it is essential to understand the relationship between variance and standard deviation. The confidence interval for the variance, denoted as \(\sigma^2\), is based on the chi-square distribution, and to find the interval for the standard deviation \(\sigma\), you take the square root of the entire variance confidence interval.

For a sample size of \(n = 16\), the degrees of freedom are \(n - 1 = 15\). Given a 90% confidence level, the significance level \(\alpha\) is \$1 - 0.90 = 0.10\(, and \)\alpha/2 = 0.05\(. Using chi-square distribution tables, the critical values for 15 degrees of freedom are approximately \)\chi^2_{\text{right}} = 25\( and \)\chi^2_{\text{left}} = 7.26\(.

The formula for the confidence interval of the population variance is:

\[\frac{(n-1) s^2}{\chi^2_{\text{right}}} < \sigma^2 < \frac{(n-1) s^2}{\chi^2_{\text{left}}}\]

To convert this into a confidence interval for the population standard deviation, take the square root of all parts:

\[\sqrt{\frac{(n-1) s^2}{\chi^2_{\text{right}}}} < \sigma < \sqrt{\frac{(n-1) s^2}{\chi^2_{\text{left}}}}\]

Here, \)s\( is the sample standard deviation, which must be squared to obtain \)s^2\(, the sample variance. For the given data, the sample standard deviation \)s\( was calculated as approximately 0.189 (in thousands of dollars), so \)s^2 = (0.189)^2$.

Plugging in the values:

\[\sqrt{\frac{15 \times (0.189)^2}{25}} < \sigma < \sqrt{\frac{15 \times (0.189)^2}{7.26}}\]

Evaluating this yields the 90% confidence interval for the population standard deviation:

\[0.146 < \sigma < 0.272\]

This interval means we are 90% confident that the true population standard deviation of daily revenue lies between 0.146 and 0.272 thousand dollars.

In a business context, this information is crucial for decision-making. For example, if the finance team expects the standard deviation of daily revenue to be no more than 0.2 thousand dollars to maintain forecast accuracy, the confidence interval suggests otherwise. Since the upper bound of the interval (0.272) exceeds 0.2, it indicates that the true standard deviation could be higher than the acceptable threshold.

Therefore, it is advisable for the analyst to recommend reviewing sales processes or forecasting models to address this potential variability. This approach ensures that the business maintains reliable revenue forecasts and can adapt strategies accordingly.

Here’s what students ask on this topic:

What is the formula for constructing a confidence interval for population variance?

To construct a confidence interval for population variance, we use the chi-square distribution. The formula for the lower bound is $(n-1)s_2/{\mathrm{\chi \sigma }}_r$ and for the upper bound is $(n-1)s_2/{\mathrm{\chi \sigma }}_l$, where $n$ is the sample size, $s_2$ is the sample variance, and ${\mathrm{\chi \sigma }}_r$ and ${\mathrm{\chi \sigma }}_l$ are the right and left critical values from the chi-square distribution with degrees of freedom $\mathrm{df}=n-\; 1$. This interval estimates the range in which the true population variance lies with a specified confidence level.

How do you find the critical values for a confidence interval for variance using the chi-square distribution?

To find the critical values for a confidence interval for variance, first determine the significance level $\alpha$ as $1\; -C$, where $C$ is the confidence level (e.g., 0.90 for 90%). Then calculate $\alpha /2$. The right critical value ${\mathrm{\chi \sigma }}_r$ corresponds to the chi-square value with area $\alpha /2$ to its right, and the left critical value ${\mathrm{\chi \sigma }}_l$ corresponds to the chi-square value with area $1\; -\alpha /2$ to its right. Use the degrees of freedom $\mathrm{df}=n-\; 1$ to look up these values in a chi-square distribution table or use statistical software.

Why is the chi-square distribution used for confidence intervals of variance instead of the normal distribution?

The chi-square distribution is used for confidence intervals of variance because the sampling distribution of the sample variance follows a chi-square distribution when the population is normally distributed. Unlike the normal distribution, the chi-square distribution is not symmetric and depends on degrees of freedom. This makes it suitable for estimating variance, which is always positive and skewed. Using the chi-square distribution allows us to find appropriate critical values to build confidence intervals for variance, which cannot be accurately done with the normal distribution.

How do you construct a confidence interval for the population standard deviation?

To construct a confidence interval for the population standard deviation, first build the confidence interval for the variance using the chi-square distribution as explained. Once you have the lower and upper bounds for variance, take the square root of both bounds. Mathematically, if the variance interval is $[L,\; U]$, then the standard deviation interval is $[\backslash sqrt\{L\},\; \backslash sqrt\{U\}]$. This method assumes the population is normally distributed and uses the same degrees of freedom $df\; =\; n\; -\; 1$ as for variance.

What assumptions must be met to use confidence intervals for population variance?

The key assumption for using confidence intervals for population variance is that the population from which the sample is drawn must be normally distributed. This is important because the derivation of the chi-square distribution for the sample variance relies on normality. Additionally, the sample should be randomly selected and independent. The degrees of freedom used in the chi-square distribution is $df\; =\; n\; -\; 1$, where $n$ is the sample size. Violating these assumptions can lead to inaccurate confidence intervals.