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:
Correct Choice:
B. There is 90% confidence that the population standard deviation is between ___ and ___.
Step-by-Step Solution:
We are asked to construct a 90% confidence interval for the population standard deviation σ\sigma, based on the sample standard deviation s=9.950s = 9.950 weeks, and a sample size of n=12n = 12.
1. Use the Chi-Square Distribution:
To find the confidence interval for the population variance σ2\sigma^2, we use the formula: ((n−1)s2χR2,(n−1)s2χL2)\left( \frac{(n – 1)s^2}{\chi^2_{R}}, \frac{(n – 1)s^2}{\chi^2_{L}} \right)
- n=12n = 12
- s=9.950s = 9.950
- Degrees of freedom: df=n−1=11df = n – 1 = 11
For a 90% confidence interval, we find the chi-square critical values:
- χL2=χ0.052\chi^2_{L} = \chi^2_{0.05} (left tail)
- χR2=χ0.952\chi^2_{R} = \chi^2_{0.95} (right tail)
From the chi-square distribution table for df = 11:
- χ0.052=19.675\chi^2_{0.05} = 19.675
- χ0.952=4.575\chi^2_{0.95} = 4.575
2. Plug Into the Variance Formula:
((11)(9.950)219.675,(11)(9.950)24.575)\left( \frac{(11)(9.950)^2}{19.675}, \frac{(11)(9.950)^2}{4.575} \right)
First, compute the numerator: (11)(9.950)2=11×99.00=1089.00(11)(9.950)^2 = 11 \times 99.00 = 1089.00
Now calculate the confidence interval for variance: (1089.0019.675,1089.004.575)=(55.331,237.997)\left( \frac{1089.00}{19.675}, \frac{1089.00}{4.575} \right) = (55.331, 237.997)
3. Convert Variance to Standard Deviation:
Take the square root of each bound:
- Lower bound: 55.331=7.437\sqrt{55.331} = 7.437
- Upper bound: 237.997=15.426\sqrt{237.997} = 15.426
✅ Final Answer:
B. There is 90% confidence that the population standard deviation is between
7.437 weeks and 15.426 weeks.
Explanation (300+ words):
This problem involves estimating the population standard deviation of the age at which babies crawl using a confidence interval derived from a sample. Confidence intervals help us determine a range of plausible values for a population parameter—in this case, the standard deviation of crawling age.
Since the data is normally distributed, the chi-square distribution is appropriate for estimating a confidence interval for the variance (and thus the standard deviation) when the population mean is unknown. The standard deviation of the sample is given as 9.950 weeks, and the sample size is 12. This gives us 11 degrees of freedom (n−1n – 1).
To create a 90% confidence interval, we use the chi-square distribution critical values that capture the central 90% of the distribution, leaving 5% in each tail. These critical values are found in the chi-square table for 11 degrees of freedom and are:
- Left-tail (5%): χ0.052=19.675\chi^2_{0.05} = 19.675
- Right-tail (95%): χ0.952=4.575\chi^2_{0.95} = 4.575
The confidence interval for variance is calculated first and then converted to standard deviation by taking the square root of both bounds. This gives the interval: (55.331,237.997)=(7.437,15.426)( \sqrt{55.331}, \sqrt{237.997} ) = (7.437, 15.426)
The correct interpretation of this result is option B: “There is 90% confidence that the population standard deviation is between 7.437 and 15.426 weeks.” This means that if we were to take many random samples and compute the interval each time, about 90% of those intervals would contain the true population standard deviation.
Option A is incorrect because confidence intervals do not express probabilities about a fixed population parameter. Option C is wrong because it misrepresents the interval as applying to sample standard deviations across repeated samples, which is not the purpose of this confidence interval.