Exercise 4 Page 629

First we will find the margin of error for the sample data. Then we will use the margin of error to find the interval, which should contain the actual percent of the population that would respond to the survey in the same way. Let's do these two things one at a time.

Margin of Error

A margin of error gives a limit on how much the responses of a sample would differ from the responses of the population. If we let n be the size of a random sample taken from a large population, we can write an expression for the margin of error. Margin of Error: ± 1/sqrt(n) We are told that in a survey of 1028 people in the U.S., 87 % reported using the Internet. Therefore, we have that n= 1028. Let's find the margin of error by substituting this value in the above expression. Margin of Error: ± 1/sqrt(1028) ≈ ± 0.031 The margin of error for the survey is about ± 0.031 or ± 3.1 %.

Interval

Now we can find the interval. Let p be the percentage of the sample responding a certain way, written as a decimal. Interval:Between p- 1sqrt(n) and p+ 1sqrt(n) We are told that 87 % of the surveyed people reported using the Internet. To find the lower boundary of the interval, we will subtract 0.031 from p= 0.87. Similarly, to find the upper boundary we will add 0.031 to p= 0.87. Let's start by finding the lower boundary.

Finally, let's find the upper boundary of the interval.

The boundaries tell us the interval in which the actual percent of the population would report using the Internet. Between 83.9 % and 90.1 %