Confidence Intervals for Population Variance: Videos & Practice Problems

Confidence intervals for variance and standard deviation use the chi-square distribution, relying on sample variance as the point estimate. Unlike confidence intervals for means, they require two critical chi-square values, $\chi l_{}$ and $\chi r_{}$, to calculate bounds: $\mathrm{Lower}=\frac{\mathrm{df}\cdot s^2}{\chi }$ and $\mathrm{Upper}=\frac{\mathrm{df}\cdot s^2}{\chi }$. For standard deviation, take square roots of these bounds. This method assumes normality and uses degrees of freedom $\mathrm{df}=n-1$. Understanding this enhances inferential statistics skills and application of chi-square distribution in variance estimation.

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 distinct critical values: one for the lower bound and one for the upper bound.

To build this interval, 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 significance level into two parts: α/2 for each tail. The critical values, χ²_l and χ²_r, correspond to the chi-square values at the cumulative probabilities of 1 - α/2 and α/2 respectively, found using chi-square distribution tables or software.

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

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

where s² is the sample variance. Notice that the larger chi-square critical value is placed in the denominator of the lower bound, resulting in a smaller lower limit, while the smaller critical value is used for the upper bound, producing a larger upper limit. This arrangement accounts for the asymmetry of the chi-square distribution.

For example, if a sample of 12 eggs yields a sample variance of 1.2 and a 90% confidence interval is desired, the degrees of freedom would be 11. The significance level α is 0.10, so α/2 = 0.05. Using chi-square tables, 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, indicating 90% confidence that the true variance 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:

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

This method assumes the underlying population is normally distributed, which is a key condition for the validity of the chi-square based confidence intervals for variance and standard deviation.

Understanding how to apply the chi-square distribution to estimate variability parameters enhances statistical inference skills, allowing for more precise conclusions about population dispersion based on sample data.

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 relationship between sample variance, degrees of freedom, and the chi-square distribution. Suppose a quality control specialist samples 30 parts from an assembly line and calculates a sample variance of 1.2 mm². Assuming the population is normally distributed, this allows for the creation of confidence intervals for the population variance.

To build a confidence interval for variance, the chi-square distribution is used with degrees of freedom equal to the sample size minus one (n - 1). For a sample size of 30, the degrees of freedom are 29. The confidence level determines the critical chi-square values, which are found using the significance level α, where α = 1 - confidence level. For a 90% confidence interval, α = 0.10, so α/2 = 0.05. The critical values correspond to the chi-square quantiles at α/2 and 1 - α/2.

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

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

where s² is the sample variance, and \(\chi^2_{\alpha/2, \, n-1}\) and \(\chi^2_{1 - \alpha/2, \, n-1}\) are the chi-square critical values for the given degrees of freedom.

For the 90% confidence interval, the critical chi-square values for 29 degrees of freedom are approximately 42.56 (right tail) and 17.71 (left tail). Plugging in the values:

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

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 52.34 (right tail) and 13.12 (left tail). Using the same formula:

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

Comparing the two intervals reveals that increasing the confidence level from 90% to 99% results in a wider confidence interval. This wider interval reflects greater uncertainty but ensures a higher probability that the true population variance is captured within the range. The lower bound decreases from 0.82 to 0.66, and the upper bound increases from 1.96 to 2.65, illustrating that higher confidence requires a broader interval.

Understanding how confidence levels affect interval width is crucial in quality control and statistical inference, as it balances precision with certainty when estimating population parameters.

Constructing Confidence Intervals for Variance or Standard Deviation Example 2

When estimating the population variance and standard deviation from a sample, especially when the data set is provided rather than summary statistics, it is essential to understand the process of constructing confidence intervals under the assumption that the population is normally distributed. For example, consider a clinical researcher measuring the systolic blood pressure of 10 patients after administering a new medication. To build a 95% confidence interval for the population variance (σ²), one must first determine the significance level α, which is 0.05 for a 95% confidence level. This leads to α/2 = 0.025 and 1 - α/2 = 0.975.

With a sample size of 10, the degrees of freedom (df) is 9. Using these values, the critical chi-square (χ²) values are found from chi-square distribution tables or statistical software: χ²_0.975,9 ≈ 2.7 and χ²_0.025,9 ≈ 19.02. The sample variance (s²) must be calculated from the raw data, often using a calculator or software. If the sample standard deviation (s) is 2.6, then the sample variance is \(s^2 = 2.6^2 = 6.76\).

The 95% confidence interval for the population variance is then computed using the formula:

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

Substituting the values:

\[\left( \frac{9 \times 6.76}{19.02}, \quad \frac{9 \times 6.76}{2.7} \right) = (3.2, \quad 22.53)\]

This interval means we are 95% confident that the true population variance lies between 3.2 and 22.53.

To find the 95% confidence interval for the population standard deviation (σ), simply take the square root of the variance interval bounds, since standard deviation is the square root of variance:

\[\left( \sqrt{3.2}, \quad \sqrt{22.53} \right) = (1.79, \quad 4.75)\]

This interval indicates that the population standard deviation is likely between 1.79 and 4.75 with 95% confidence.

Understanding how to calculate confidence intervals for variance and standard deviation using chi-square distributions is crucial in inferential statistics, especially when working with normally distributed populations. It involves identifying the correct critical values, calculating sample variance from raw data, and applying the formulas accurately to interpret the variability within a population based on sample data.

Here’s what students ask on this topic:

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

To construct a confidence interval for a population variance, you start with the sample variance as your point estimate. Since the chi-square distribution is asymmetrical, you use two critical values, ${\mathrm{\backslash chi}}_r²$ and ${\mathrm{\backslash chi}}_l²$, instead of a margin of error. These critical values are found using the degrees of freedom, which is $n-1$, and the confidence level. The lower bound of the interval is calculated as $\frac{\mathrm{df}\times s^2}{{\mathrm{\backslash chi}}_r}$, and the upper bound as $\frac{\mathrm{df}\times s^2}{{\mathrm{\backslash chi}}_l}$. This interval estimates the range where the true population variance lies with the specified confidence level. It is important that the population is normally distributed for this method to be valid.

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

To find the critical values for a confidence interval of variance, you first determine the significance level $\mathrm{\backslash alpha}$ as $1-C$, where $C$ is the confidence level. Then calculate $\frac{\mathrm{\backslash alpha}}{2}$ and its complement $1-\frac{\mathrm{\backslash alpha}}{2}$. Using the degrees of freedom $\mathrm{df}=n-1$, look up these two probabilities in the chi-square distribution table to find the right critical value ${\mathrm{\backslash chi}}_r²$ corresponding to $\frac{\mathrm{\backslash alpha}}{2}$ and the left critical value ${\mathrm{\backslash chi}}_l²$ corresponding to $1-\frac{\mathrm{\backslash alpha}}{2}$. These values are then used to calculate the confidence interval bounds for the variance.

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 asymmetrical and depends on degrees of freedom. This asymmetry means we cannot use a simple margin of error as with means; instead, we use two critical values from the chi-square distribution to form the interval. This approach accurately reflects the variability and skewness in the distribution of sample variances, making it the appropriate choice for estimating population variance.

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

To construct a confidence interval for the population standard deviation, first calculate the confidence interval for the variance using the chi-square distribution as usual. Once you have the lower and upper bounds for the 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 works because the standard deviation is the square root of the variance. This method assumes the population is normally distributed and uses the same critical values and degrees of freedom as the variance interval.

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

The key assumption for constructing confidence intervals for population variance is that the population from which the sample is drawn must be normally distributed. This assumption is crucial because the derivation of the chi-square distribution for the sample variance relies on normality. If the population is not normal, the chi-square distribution may not accurately describe the sampling distribution of the variance, leading to invalid confidence intervals. Additionally, the sample should be randomly selected and independent. Meeting these assumptions ensures the confidence interval provides a reliable estimate of the true population variance.