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Literate Adults. Suppose that you have been hired to estimate the percentage of adults in your state who are literate. You take a random sample of 100adults and find that 96are literate. You then obtain a 95%confidence interval of

0.96±1.96·(0.96)(0.04)/100

or 0.922to 0.998. From it you conclude that you can be 95%confident that the percentage of all adults in your state who are literate is somewhere between 92.2%and 99.8%. Is anything wrong with this reasoning?

Short Answer

Expert verified

The procedure is not appropriate.

Step by step solution

01

Given Information

A random sample of 100adults and find that 96are literate then obtain a 95%confidence interval of

0.96±0.96·(0.96)(0.04)/100

02

Explanation

Yes, the procedure is not appropriate.

Here x=96

and n=100

n-x=100-96

=4

Both xand n-xare both 5the one-proportion z-interval technique isn't appropriate if the number of proportions is larger than one.

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