Question

A simple random sample from a population with a normal distribution of 107 body temperatures has x = 98.30°F and S=0.61°F. Construct a 99% confidence interval estimate of the standard deviation of body temperature of all healthy humans. Click the icon to view the table of Chi-Square critical values. 1°F<o< °F (Round to two decimal places as needed.)

Answer 1

Answer: given n=107, 7= 98.30r, s= 0.61 To construct the 99%. C. I, the formula is (n-1) x² Using chi-square distribution table with df=n-1= 107-1 = 106, alpha = 1-0.99 = 0.01 +2 * - 1n-1 X-0.01, 106 = x 995,106 = 22:25 * 1 1 1 -1 = X 0 0", 106 = *00051 106 = 147.247 +2 < o L (107-1) x (0.612 147.247 (107-1) + (0.612 I 72.251 106x0.3721 1064 0.3721 72.251 1471247 0.26 79 LoL 0.5459

0.5176 5 L 0.7389 lower limit 2 0.5176 upper limit = 0.7389 So, 99% col for standard deviation is [0.5176 PFLOL [0.7389] F NOTE! Please like please like cee