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A clinical trial was conducted to test the effectiveness of a drug treating insomnia in older subjects. After the treatment with the drug, 28 subjects had a mean wake time of 97.9 minutes with a standard deviation of 41.2 minutes. Assume that the 28 sample values appear to be from a normally distributed population. Construct a 95% confidence interval estimate of the standard deviation of wake times for a population with drug treatments.

Answer :

Assuming the sample values are normally distributed, we can construct a 95% confidence interval estimate of the standard deviation of wake times for the population with drug treatments.

Using the formula for a confidence interval estimate of a population standard deviation, we can calculate the lower and upper bounds of the interval. The formula is:

(Lower Bound) ≤ σ ≤ (Upper Bound)

Where:
Lower Bound = √(n - 1) x s^2 / χ^2 (α/2, n-1)
Upper Bound = √(n - 1) x s^2 / χ^2 (1-α/2, n-1)
n = sample size (28)
s = sample standard deviation (41.2 minutes)
χ^2 (α/2, n-1) = chi-square value for α/2 and n-1 degrees of freedom (from chi-square distribution table)
χ^2 (1-α/2, n-1) = chi-square value for 1-α/2 and n-1 degrees of freedom (from chi-square distribution table)
α = level of significance (0.05 for a 95% confidence interval)
Plugging in the values, we get:
Lower Bound = √(28-1) x (41.2)^2 / χ^2 (0.025, 27) = 26.6
Upper Bound = √(28-1) x (41.2)^2 / χ^2 (0.975, 27) = 65.6
Therefore, the 95% confidence interval estimate of the standard deviation of wake times for a population with drug treatments is between 26.6 and 65.6 minutes. This means that we are 95% confident that the true population standard deviation of wake times for older subjects treated with the drug for insomnia falls within this range.
To construct a 95% confidence interval estimate of the standard deviation of wake times for a population with drug treatments, follow these steps:

1. Identify the sample size (n), sample standard deviation (s), and the chi-square values from the chi-square distribution table for the given confidence level:
- Sample size (n): 28 subjects
- Sample standard deviation (s): 41.2 minutes

For a 95% confidence level with 27 degrees of freedom (n-1), the chi-square values are:
- Lower chi-square value (χ²₁): 14.573
- Upper chi-square value (χ²₂): 41.337

2. Apply the chi-square formula to calculate the lower and upper limits of the confidence interval:
Lower limit = √((n - 1) × s² / χ²₂) = √((28 - 1) × 41.2² / 41.337) = 31.1 minutes
Upper limit = √((n - 1) × s² / χ²₁) = √((28 - 1) × 41.2² / 14.573) = 62.0 minutes

3. Interpret the result:
The 95% confidence interval estimate for the standard deviation of wake times in a population treated with the drug for insomnia is between 31.1 minutes and 62.0 minutes. This means that we are 95% confident that the true standard deviation of wake times for this population lies within this range.

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