Question

A graduate student wishes to know the proportion of U.S. adults who speak two or more languages. He surveys 428 U.S. adults and finds that 107 speak two or more languages. Construct a 99% confidence interval to estimate the proportion of all U.S. adults that speak two or more languages.

Answer 1

Answer)

First we need to estimate point of estimate

P = 107/428

As 107 speak two or more languages out of 428

P = 107/428 = 0.25

Now we need to determine the margin of error

= z*√{(p*(1-p))/n}

Here n = 428

We will solve this step by step p*(1-p) = 0.25*(1-0.25)

= 0.1875

Now we will divide it by n which is 428

0.00043808411

Now we will take the under root of it

= 0.02093045895

Now we will multiply it with z value of 99% confidence interval which is 2.58

2.58*0.0093045895

= 0.05400058410, this is our margin of error

Confidence interval is

P-margin of error, P+margin of error

0.25 - 0.05400058410, 0.25+0.05400058410

= 0.290005841, 0.790005841