Which of the following changes in sample size and degrees of freedom, respectively, is MOST likely to result in an increase in the width of the confidence interval for the population mean? Sample Size Deg of Freedom a. increase increase b. increase decrease c. decrease increase d. decrease decrease I incorrectly chose A… Can someone please help me understand this concept? If you have the 2007 cfai curriculum, what pages can I read for his problem? Thank you!! Will post the correct answer eventually.

So if a C.I. for the mean is given by (X-bar - t*s/Sqrt(n), X-bar + t*s/Sqrt(n)) then the width of the C.I. is 2*t*s/Sqrt(n) The only fact that you need to know besides that is that for any significance level t decreases as d.f. increase (look on a table). So as d.f. decreases, t increases, so width increases. As n decreases the width obviously goes up. The way to think about this is that as n increases you have more information in X-bar for estimating mu so your C.I. width should decrease. As d.f. go up, the error in using s instead of sigma goes down because you have more information for estimating sigma so C.I. width should decrease.

Is it D? Just guessing. But with smaller sample size the probability to match the real mean decrease und the confidence interval becomes wider. And with higher degrees of freedom we approach to normal distribution.

sx/sqrt(n) is one of the components as n increases sx/sqrt(n) decreases and since width of confidence interval = xbar +/- ref value * sx/sqrt(n) the confidence interval will increase when n decreases. so choice is either c or d as n decreases, dof (typically n-1) also decreases as dof decreases, t-value increases t 0.025, 30 = 2.042 t 0.025, 29 = 2.045 and increasing the number - also increases the width of the confidence interval. So D?

Oh yeah - The answer is D.

Thanks guys, yeah answer is D.