Question 1193340
Because b has more information, its interval will be narrower. Ci also is inversely proportional to the square root of the sample size and directly proportional to the t-value which is also less for a larger n, everything else's being the same.
95% half-interval of a is t(0.975,df=10)*9.49/sqrt(11); the 9.49 is the sd, the sqrt(90), and the units are $.
=2.228*9.49/sqrt(11)=6.375 or 6.38
the interval is ($393.62, $406.38)
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For the second, t is 2.101, and half-interval is 2.101*9.49/sqrt(19)=4.57
interval is ($395.43, $404.57) It is narrower