Question

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation o. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. College students' annual earnings: 98% confidence; n = 9, sample mean is $4121, s = $873 $551<o< $1925 O $590 < 0 < $1672 $531 < o < $1708 $687 <o< $1139

Answer 1

n= 9 Solution ã - $ 4121, 5= $873 98% confidence for standard deviation () = ? (n-1)52 sos (n-152 of is the (1-x)% دابه xi-a2 chisquare we have confidence interval for o with (hr) degree of freedom n-1=8 degree of freedom for a 98% Confidence internal d=0.02, which gives 1 to of the area at each end of the chi square distribution from 2² distribution table =10646 and X o of 13• 362 arva 16.49 2 98% confidence interval for a 8x (873)2 8X(873) so 13362 1.646 넌 675 , 1925) So Answer (675404 1925) CS Scanned with CamScanner