Confidence Intervals For Proportions and 

Choosing the Sample Size

A Large Sample Confidence Interval for a Population Proportion

Recall that a confidence interval for a population mean is given by

Confidence Interval for a Population Mean                                          zc s                           x

We can make a similar construction for a confidence interval for a population proportion.  Instead of x, we can use p and instead of s, we use , hence, we can write the confidence interval for a large sample proportion as

Confidence Interval Margin of Error for a Population Proportion

Example

1000 randomly selected Americans were asked if they believed the minimum wage should be raised.  600 said yes.  Construct a 95% confidence interval for the proportion of Americans who believe that the minimum wage should be raised.

Solution:  

We have 

        p = 600/1000 = .6         z_c = 1.96         and     n = 1000  

We calculate:

Hence we can conclude that between 57 and 63 percent of all Americans agree with the proposal. In other words, with a margin of error of .03 , 60% agree.

Calculating n for Estimating a Mean

Example

Suppose that you were interested in the average number of units that students take at a two year college to get an AA degree.  Suppose you wanted to find a 95% confidence interval with a margin of error of  .5 for m knowing s = 10.  How many people should we ask?

Solution

Solving for n in

        Margin of Error  =  E  = z_c s/

we have

        E =  z_cs

                          z_c s  
          =                     
                            E

Squaring both sides, we get

We use the formula:  

Example

A Subaru dealer wants to find out the age of their customers  (for advertising purposes).  They want the margin of error to be 3 years old.  If they want a 90% confidence interval, how many people do they need to know about?

Solution: 

We have 

        E = 3,       z_c = 1.65

but there is no way of finding sigma exactly.  They use the following reasoning: most car customers are between 16 and 68 years old hence the range is 

        Range  =  68 - 16  =  52

The range covers about four standard deviations hence one standard deviation is about

        s  @  52/4  =  13

We can now calculate n:

Hence the dealer should survey at least 52 people.

Finding n to Estimate a Proportion

Example 

Suppose that you are in charge to see if dropping a computer will damage it.  You want to find the proportion of computers that break.  If you want a 90% confidence interval for this proportion, with a margin of error of   4%, How many computers should you drop?

Solution

The formula states that

Squaring both sides, we get that

                      z_c² p(1 - p)
          E²  =                              
                             n

Multiplying by n, we get

        nE²  =  z_c²[p(1 - p)]

This is the formula for finding n.

Since we do not know p, we use .5  ( A conservative estimate)

          We round 425.4 up for greater accuracy

We will need to drop at least 426 computers.  This could get expensive.

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