The following $95\%$ confidence interval was constructed using a large sample of data: $(86.52,89.48)$. Which of the following could be a $99\%$ confidence interval for the same set of data?

$I. (86.98,89.02)$

$II. (86.37,89.63)$

$III. (87.04,88.98)$

My attempt: It is a large sample of data so we can approximate the sampling distribution with a Normal model. The mean is $$\bar{x} = \frac{86.52+89.48}{2}=88$$ The margin of error for the $95\%$ confidence interval is $$z^*\cdot (\text{Standard Error}) = 1.96(SE) = (89.48-88) = 1.48$$ This gives us that $SE$ is $.755$. The critical $z$ value for a $99\%$ interval is about $2.58$. The new margin of error is now $.755\cdot2.58 = 1.95$ So the $99\%$ confidence interval is now $(88-1.95,88+1.95) = (86.05,89.95)$. Which is not one of the answers. Where did I go wrong?

  • 1
    $\begingroup$ The question only says "could be", not "is", maybe you were not expected to do any calculations. $\endgroup$
    – David
    Mar 11 '16 at 4:02

A higher confidence level will merely widen the interval. This leaves choice $II.$

  • $\begingroup$ Yes but why does my method not give the exact answer? $\endgroup$ Mar 11 '16 at 4:01
  • $\begingroup$ @JamesCooper I feel that the question was designed to be more theoretical, rather than computational. $\endgroup$
    – Chris
    Mar 11 '16 at 4:08
  • $\begingroup$ @james because you have no idea what is the distribution of the underlying parameter. Remember, most thing in life aren't normal. $\endgroup$
    – A.S.
    Mar 11 '16 at 4:11

It is the mean of any distribution that is asymptotically normal. Your 'large sample of data' isn't necessarily normal -- it could follow any distribution. Since a wider CI is always wider, it must be II.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.