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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?

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

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  • $\begingroup$ Yes but why does my method not give the exact answer? $\endgroup$ – James Cooper 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
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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.

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