Question

10.1.4-T Question Help Assume that you have a sample of n, = 9, with the sample mean X, = 42, and a sample standard deviation of S, = 7, and you have an independent sample of n = 13 from another population with a sample mean of X, = 37, and the sample standard deviation Sy = 8. Construct a 99% confidence interval estimate of the population mean difference between 11 and 12. Assume that the two population variances are equal. $11-H2=0 (Round to two decimal places as needed.) Enter your answer in the edit fields and then click Check Answer. All parts showing Clear All Check Answer

Answer 1

Confidence interval is :-

( X̅1 - X̅2 ) ± t( α/2 , n1+n2-2) SP √( (1/n1) + (1/n2))
t(α/2, n1 + n1 - 2) = t( 0.01/2, 9 + 13 - 2) = 2.845

Sp= ((n1 – 1).S+ + (n2 – 1)S3)/(n1+ n2 – 2)


Sp = (V(9 - 1)72 + (13 – 1)82)/(9+ 13 – 2)


Sp = 7.6158

( 42 - 37 ) ± t(0.01/2 , 9 + 13 -2) 7.6158 √ ( (1/9) + (1/13))
Lower Limit = ( 42 - 37 ) - t(0.01/2 , 9 + 13 -2) 7.6158 √( (1/9) + (1/13))
Lower Limit = -4.40
Upper Limit = ( 42 - 37 ) + t(0.01/2 , 9 + 13 -2) 7.6158 √( (1/9) + (1/13))
Upper Limit = 14.40
99% Confidence Interval is ( -4.40 , 14.40 ).