Pls I need help with this 2 questions, I\'m not sure if I\'m my answer is correc
ID: 3222071 • Letter: P
Question
Pls I need help with this 2 questions, I'm not sure if I'm my answer is correct. Pls show your work. ThanksMoving to another question wll save this response Question 4 of 8 Question 4 10 points Saved Suppose two independent random samples of size n1 5 and n2 9 are taken from two independent normal populations, respectively The computod sample standard are s 4 and s2 7. respectvely A 98% confidence interval for the rato, R orst population standard deviation/second populawon standard deviaton), is a 2069, 21 4592) b 11 2198, 2 1012l c lo 2158, 2 1983 d lo 4540, 4 6339 A Moving to another queston save this response Ques hon 4 of 8
Explanation / Answer
Q.4 confidence interval for the ratio of standard deviation.
R = first population std. dev./ second population std. dev. = 4/7 = 0.5714
98% confidence interval
Lower confidence limit = (sx/ sy) sqrt[ 1/ F/2, (n-1,m-1) ]
=(sx/ sy) sqrt[ 1/ F/2, (5-1,9-1) ]
= (4/7) sqrt [ 1/ F0.01, (4,8) ]
= (4/7) sqrt[1/7.006] = 0.2158
Upper confidence limit = (sx/ sy) sqrt[ 1/ F/2, (m-1,n-1) ]
= (sx/ sy) sqrt[ F/2, (9-1,5-1) ] =
= (4/7) sqrt [ F0.01, (8,4) ]
= (4/7) sqrt[14.799] = 2.1983
option C is correct.
Q.3 99% confidence Interval ; dF = (8-1) + (12-1) = 18
so tcritical = 2.878
=(x1 -bar - x2- bar) +- 2.878 sqrt{ [ (n1 -1) * s12) + (n2 -1) * s22) ] / (n1 + n2 - 2) * (1/n1 + 1/n2)}
this formula is used whe n1 and n2 is less than 30 and population variance is unknown.
= (2.20 - 2.29) +- sqrt {[7 * (0.11)2 + 11 * (0.07)2)]/18 * ( 1/8 + 1/12)}