Math Problem Statement

The following data represent the pH of rain for a random sample of 12 rain dates in a particular region. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. The sample standard deviation is sequals0.332. Construct and interpret a 99​% confidence interval for the standard deviation pH of rainwater in this region. 4.66 5.22 5.10 4.71 4.77 4.86 5.82 4.85 5.10 4.84 4.76 4.64

Solution

To construct a 99% confidence interval for the standard deviation of the pH of rainwater, we will use the chi-square distribution because the data is assumed to be normally distributed, and we are working with a sample standard deviation.

Given:

  • Sample size n=12n = 12
  • Sample standard deviation s=0.332s = 0.332
  • pH values:
    4.66, 5.22, 5.10, 4.71, 4.77, 4.86, 5.82, 4.85, 5.10, 4.84, 4.76, 4.64

Steps to Calculate the Confidence Interval for the Standard Deviation:

  1. Calculate the sample variance: s2=(0.332)2=0.110224s^2 = (0.332)^2 = 0.110224

  2. Degrees of freedom: df=n1=121=11df = n - 1 = 12 - 1 = 11

  3. Determine the chi-square critical values for a 99% confidence interval. Using the chi-square distribution table (or a calculator), we find the chi-square critical values for 11 degrees of freedom at the 0.5% and 99.5% percentiles (since we want 99% confidence, we look at the outer 0.5% tails):

    • χ0.005,1123.053\chi^2_{0.005,11} \approx 3.053
    • χ0.995,11226.757\chi^2_{0.995,11} \approx 26.757
  4. Compute the confidence interval for the population variance σ2\sigma^2: (n1)s2χ0.9952σ2(n1)s2χ0.0052\frac{(n - 1) s^2}{\chi^2_{0.995}} \leq \sigma^2 \leq \frac{(n - 1) s^2}{\chi^2_{0.005}} Substituting the values: 11×0.11022426.757σ211×0.1102243.053\frac{11 \times 0.110224}{26.757} \leq \sigma^2 \leq \frac{11 \times 0.110224}{3.053} 0.0453σ20.39760.0453 \leq \sigma^2 \leq 0.3976

  5. Take the square root to find the confidence interval for the population standard deviation σ\sigma: 0.0453σ0.3976\sqrt{0.0453} \leq \sigma \leq \sqrt{0.3976} 0.213σ0.6310.213 \leq \sigma \leq 0.631

Interpretation:

We are 99% confident that the true population standard deviation of the pH of rainwater in this region lies between 0.213 and 0.631.


Would you like further details or have any questions on this process?

Here are 5 questions for you to explore based on this:

  1. How do changes in sample size affect the width of the confidence interval?
  2. What assumptions must hold for the chi-square distribution to be applicable in this context?
  3. How would the interval change if the confidence level were reduced to 95%?
  4. Can you compare this interval with one for the mean and discuss the differences?
  5. What role do outliers play in computing confidence intervals for standard deviation?

Tip: Increasing the confidence level widens the confidence interval but also increases certainty in capturing the true parameter.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Chi-Square Distribution

Formulas

Sample variance: s^2 = (sample standard deviation)^2
Chi-square confidence interval for population variance: (n-1)s^2 / χ²_(α/2, df) ≤ σ² ≤ (n-1)s^2 / χ²_(1-α/2, df)
Population standard deviation: σ = √(σ²)

Theorems

Chi-Square Distribution

Suitable Grade Level

Grades 11-12, College Level