llme Remaining: 00:59:32 Submit Te This Question: 1 pt 16 of 22 (5 complete) Thi
ID: 2932646 • Letter: L
Question
llme Remaining: 00:59:32 Submit Te This Question: 1 pt 16 of 22 (5 complete) This Test: 22 pts possi Question Help A simple random sample from a population with a normal distr bution o 9 body temperatures has x-9870 and s-o 66 F Construct an 80% confidence estimate of the standard deviation of body temperature of all healthy humans terval Click the icon to view the table of Chi-Square critic al values Round to two decimal places as needed) Enter your answer in each of the answer boxes 8.1 10/28/17 11:59pm 8.2Explanation / Answer
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s = standard deviation
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since alpha =0.2
^2 right = (1 - confidence level)/2 = (1 - 0.8)/2 = 0.2/2 = 0.1
^2 left = 1 - ^2 right = 1 - 0.1 = 0.9
the two critical values ^2 left, ^2 right at 98 df are 116.32 , 80.541
s.d( s^2 )=0.66
sample size(n)=99
confidence interval for ^2= [ 98 * 0.44/116.32 < ^2 < 98 * 0.44/80.541 ]
= [ 42.69/116.32 < ^2 < 42.69/80.54 ]
[ 0.37 < ^2 < 0.53 ]
and confidence interval for = sqrt(lower) < < sqrt(upper)
= [ sqrt (0.37) < < sqrt(0.53), ]
= [ 0.61 < < 0.73 ]
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