Test: Test 1 02:30 00 Sint T This Question: 10 pts 1301 27(17 complete) ? This T

Question

Test: Test 1 02:30 00 Sint T This Question: 10 pts 1301 27(17 complete) ? This Test: 100 pts poss Use the given information to tind the number of degrees of freedom, the cntical vales xi and xi and the conlidnce interval estimate of o It is masonsaible to assume that a simple sardkm sample hes b population with a normal distribution a Nicotro in menthol cigarettes 98% corlickince n=21.3-025mg Sque otical va Cick the icon to view the table of d-(lype a whole number) Round to threse dacimad places as ndd) x(Round to tteee decimal places as weeded ) The confidence interval essimate of o is mgc am Round to hwo decimal places as needed) Enter your answer in each of the answer boxes

Explanation / Answer

Solution:- Given that n =21, s = 0.25mg confidence interval 98%

df = n-1 = 21-1 = 20

XL^2 = 8.260

XR^2 = 37.570

The confidence interval for the ?
: sqrt(((n-1)*s^2)/XR^2) < ? < sqrt(((n-1)*s^2)/XL^2)

: sqrt((20*0.25^2)/37.570) < ? < sqrt((20*0.25^2)/8.260)

: 0.18 < ? < 0.39

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if the confidence interval for the 90%

df = n-1 = 21-1 = 20

XL^2 = 10.850

XR^2 = 31.410

The confidence interval for the ?
: sqrt(((n-1)*s^2)/XR^2) < ? < sqrt(((n-1)*s^2)/XL^2)

: sqrt((20*0.25^2)/31.410) < ? < sqrt((20*0.25^2)/10.850)

: 0.20 <  ? < 0.34

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Note : the confidence not clear ...

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