# Confidence Intervals for a Large Sample

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?

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

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