Question

You randomly select and measure the contents of 10 bottles of cough syrup. The results (in fluid ounces) are shown to the right. 4.291 4.244 4.214 4.259 4.188 4.267 4.247 4.224 4.266 4.233 Assume the sample is taken from a normally distributed population. Construct 99% confidence intervals for (a) the population variance o2 and (b) the population standard deviation o. Interpret the results ). (a) The confidence interval for the population variance is (Round to six decimal places as needed.) Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to six decimal place as needed.) A. With 1% confidence, it can be said that the population variance is between B. With 99% confidence, it can be said that the population variance is between and and C. With 99% confidence, it can be said that the population variance is greater than D. With 1% confidence, it can be said that the population variance is less than (b) The confidence interval for the population standard deviation is (Round to four decimal places as needed.) Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to four decimal place as needed.) A. With 1% confidence, you can say that the population standard deviation is between inches B. With 99% confidence, you can say that the population standard deviation is between and and inches. C. With 1% confidence, you can say that the population standard deviation is greater than D. With 99% confidence, you can say that the population standard deviation is less than inches inches. O O

Answer 1

First, we have to find the sample variance and standard deviation

4.214 17.757796 4.291 18.412681 4.259 18.139081 4.244 18.011536 4.188 17.539344 4.267 18.207289 4.266 18.198756 4.247 18.037009 4.224 17.842176 4.233 17.918289 Sum = 42.433 180.064

The sample mean X is computed as follows: * = x = 42,439 – 4,25 Also, the sample variance s2 is -= (3x- : (*)) - 1–(1800 ,)

(upto six decimal place)

S2 = 0,000890

a)

The critical values for a = 0.01 and df = 9 degrees of freedom are xî = xỉ-/2,-1 = 1.7349 and xi = x2-1 = 23.5894. The corresponding confidence interval is computed as shown below: CI(Variance) = ((n − 1)g?, (n − 1)52 ) ( x 2,1-1 xi-a 2,4-1,

CI = (10 - 1)(0.000890) (10 - 1)(0.000890) 23.5894 1.7349

The confidence interval for the population variance is

CI = (0.000340,0.004617)

for interpretation

option B is right

With 99% confidence, it can be said that the population variance is between 0.000340 and 0.004617

b) Now that we have calculated the limits for the confidence interval, the limits for the 99% confidence interval for the population standard deviation can be obtained by simply taking the square root of the limits of CI for the variance, so then:

The confidence interval for the population standard deviation is

CI = (10.000340, 0.004617)

CI = (0.0184, 0.0679)

for interpretation

option B is right

With 99% confidence, it can be said that the population standard deviation is between 0.0184 and 0.0679