Question

Math 120/12 Statistics w/ Corequisite Spr 20 (1:30) Stephanie Aguirre : 04/13/20 10:01 PM Homework: Section 9.3 Save Score: 0 of 3 pts 6 of 6 (6 complete) HW Score: 85%, 17 of 20 pts X 9.3.14 Question Help 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 = 10.157 weeks. Construct and interpret a 99% confidence interval for the population standard deviation of the age (in weeks) at which babies first crawl. 53 29 44 37 38 26 2 48 37 56 26 38 29 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.) O A. If repeated samples are taken, 99% of them will have the sample standard deviation between and O B. There is a 99% probability that the true population standard deviation is between and . O C . There is 99% confidence that the population standard deviation is between and

Answer 1

The interpretation for standard deviation is given in option c)

We need to construct the 99% confidence interval for the population variance. We have been provided with the following information about the sample variance and sample size:

s^2 =	103.164649
n =	12

The critical values for α=0.01 and df = 11 degrees of freedom are$\chi_L^2 = \chi^2_{1-\alpha/2,n-1} = 2.6032$and $\chi_U^2 = \chi^2_{\alpha/2,n-1} = 26.7568$ . The corresponding confidence interval is computed as shown below:

$CI(\text{Variance})= \displaystyle \left( \frac{(n-1) s^2}{\chi^2_{\alpha/2,n-1}}, \frac{(n-1) s^2}{\chi^2_{1-\alpha/2,n-1} } \right)$

$CI(\text{Variance})= \displaystyle \left( \frac{(12-1) \times 103.1646}{26.7568}, \frac{(12-1) \times 103.1646}{2.6032} \right)$

$CI(\text{Variance})= (42.412, 435.9256)$

Now that we have the limits for the confidence interval, the limits for the 99% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then:

$CI(\text{Standard Deviation}) = \displaystyle \left( \sqrt{42.412} , \sqrt{435.9256} \right) = \left( 6.5124 , 20.8788\right)$

c) There is 99% confidence that the population standard deviation is betwen 6.5124 and 20.8788