The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers. The data are normally distributed and s = 9.950 weeks. Construct and interpret a 90% confidence interval for the population standard deviation of the age (in weeks) at which babies first crawl. Click the icon to view the table of critical values of the chi-square distribution. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to three decimal places as needed.)
A. There is a 90% probability that the true population standard deviation is between and
B. There is 90% confidence that the population standard deviation is between and
C. If repeated samples are taken, 90% of them will have the sample standard deviation between and
The Correct Answer and Explanation is:
We are asked to construct a 90% confidence interval for the population standard deviation using sample data. Here’s how to do it step-by-step.
Given Information:
- Sample size ($n$) = 12
- Sample standard deviation ($s$) = 9.950 weeks
- The data is normally distributed
- Confidence level = 90%
- Distribution used: Chi-square distribution
- Degrees of freedom ($df$) = $n – 1 = 11$
Step 1: Use Chi-Square Formula for Confidence Interval
To find the confidence interval for the population standard deviation $\sigma$, use the formula:
$$
\left( \sqrt{\frac{(n – 1) s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n – 1) s^2}{\chi^2_{1 – \alpha/2}}} \right)
$$
Where:
- $\chi^2_{\alpha/2}$ and $\chi^2_{1 – \alpha/2}$ are critical values from the chi-square distribution
- $\alpha = 1 – 0.90 = 0.10 \Rightarrow \alpha/2 = 0.05$
Step 2: Find Chi-Square Critical Values
From the chi-square table for $df = 11$:
- $\chi^2_{0.05} = 19.675$
- $\chi^2_{0.95} = 4.575$
Step 3: Plug into the Formula
$$
\text{Lower bound} = \sqrt{\frac{(11)(9.950)^2}{19.675}} = \sqrt{\frac{(11)(99.00)}{19.675}} = \sqrt{\frac{1089.96}{19.675}} = \sqrt{55.39} \approx 7.441
$$
$$
\text{Upper bound} = \sqrt{\frac{(11)(9.950)^2}{4.575}} = \sqrt{\frac{1089.96}{4.575}} = \sqrt{238.16} \approx 15.433
$$
Answer:
$$
\boxed{(7.441,\ 15.433)}
$$
So the correct choice is:
B. There is 90% confidence that the population standard deviation is between 7.441 and 15.433 weeks.
Explanation (300+ words):
In inferential statistics, confidence intervals allow us to estimate a population parameter (like the standard deviation) based on a sample. In this case, we aim to estimate the population standard deviation of the age at which babies first crawl using a 90% confidence interval.
We start with a sample standard deviation (s = 9.950 weeks) taken from 12 babies. Since the sample size is small and the population is normally distributed, we use the chi-square distribution to construct the confidence interval for the population standard deviation.
The formula for the confidence interval of a population standard deviation relies on chi-square critical values and the sample variance. We determine the degrees of freedom (df = n – 1 = 11) and look up the chi-square values corresponding to the 5th and 95th percentiles (because it’s a 90% confidence interval, and the remaining 10% is split between the two tails of the distribution).
Using these chi-square values (4.575 and 19.675), we plug them into the standard deviation formula to get the interval:
- Lower limit ≈ 7.441 weeks
- Upper limit ≈ 15.433 weeks
This tells us that we can be 90% confident the true population standard deviation of the age when babies first crawl lies within this range. This does not mean there’s a 90% chance the true value is in this interval—it means that if we repeated this sampling many times, 90% of the calculated intervals would contain the true standard deviation.
Option B is the correct interpretation, while A misuses the word “probability” (which applies before sampling), and C incorrectly describes the behavior of sample standard deviations, not the confidence interval for the population standard deviation.
✅ Final Answer:
B. There is 90% confidence that the population standard deviation is between 7.441 and 15.433 weeks.