Math Problem Statement

Construct the confidence interval for the population standard deviation for the given values. Round your answers to one decimal place. n = 20, s = 3.8, and c = 0.8

Solution

To construct the confidence interval for the population standard deviation based on the sample standard deviation, we need to use the Chi-square distribution. Given the parameters:

  • n=20n = 20 (sample size)
  • s=3.8s = 3.8 (sample standard deviation)
  • c=0.8c = 0.8 (confidence level)

Step 1: Determine Degrees of Freedom

The degrees of freedom df=n1=201=19df = n - 1 = 20 - 1 = 19.

Step 2: Find Chi-Square Critical Values

For a confidence level c=0.8c = 0.8, we want the middle 80% of the chi-square distribution, which leaves 10% in each tail. Therefore:

  • The lower critical value χα/22\chi^2_{\alpha/2} is found at the 0.10 percentile.
  • The upper critical value χ1α/22\chi^2_{1 - \alpha/2} is found at the 0.90 percentile.

Using a chi-square table or calculator for 19 degrees of freedom:

  • χ0.10,19210.117\chi^2_{0.10, 19} \approx 10.117
  • χ0.90,19230.144\chi^2_{0.90, 19} \approx 30.144

Step 3: Calculate the Confidence Interval

The confidence interval for the population standard deviation σ\sigma is given by: (n1)s2χ1α/22<σ<(n1)s2χα/22\sqrt{\frac{(n - 1) s^2}{\chi^2_{1 - \alpha/2}}} < \sigma < \sqrt{\frac{(n - 1) s^2}{\chi^2_{\alpha/2}}}

Substitute the values: (201)(3.8)230.144<σ<(201)(3.8)210.117\sqrt{\frac{(20 - 1) \cdot (3.8)^2}{30.144}} < \sigma < \sqrt{\frac{(20 - 1) \cdot (3.8)^2}{10.117}}

1914.4430.144<σ<1914.4410.117\sqrt{\frac{19 \cdot 14.44}{30.144}} < \sigma < \sqrt{\frac{19 \cdot 14.44}{10.117}}

Calculating each side:

  • Lower bound: 274.3630.1449.13.0\sqrt{\frac{274.36}{30.144}} \approx \sqrt{9.1} \approx 3.0
  • Upper bound: 274.3610.11727.125.2\sqrt{\frac{274.36}{10.117}} \approx \sqrt{27.12} \approx 5.2

Answer:

The 80% confidence interval for the population standard deviation σ\sigma is approximately: 3.0<σ<5.23.0 < \sigma < 5.2

Would you like further explanation on any part of this process?

Here are some related questions for further exploration:

  1. How does changing the confidence level affect the width of the confidence interval?
  2. What would the confidence interval be if the sample size increased?
  3. Why is the chi-square distribution used for confidence intervals of standard deviation?
  4. How do we interpret the confidence interval in a practical context?
  5. What assumptions are required for this interval to be valid?

Tip: When using chi-square tables, ensure you choose the correct degrees of freedom and the appropriate percentiles for your confidence level.

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Math Problem Analysis

Mathematical Concepts

Confidence Interval
Chi-square Distribution
Population Standard Deviation

Formulas

Degrees of Freedom: df = n - 1
Confidence Interval Formula: sqrt((n - 1) * s^2 / chi-square upper bound) < σ < sqrt((n - 1) * s^2 / chi-square lower bound)

Theorems

Chi-square Distribution

Suitable Grade Level

College level or advanced high school statistics