Question

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 $100$adults and find that $96$are literate. You then obtain a $95\%$confidence interval of

$0.96\pm 1.96·\sqrt{(0.96)(0.04)/100}$

or $0.922$to $0.998$. 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

The procedure is not appropriate.

Step-by-step solution

Given Information

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

$0.96\pm 0.96·\sqrt{(0.96)(0.04)/100}$

Explanation

Yes, the procedure is not appropriate.

Here $x=96$

and $n=100$

$n-x=100-96$

$=4$

Both $x$and $n-x$are both $5$the one-proportion z-interval technique isn't appropriate if the number of proportions is larger than one.