Confidence Intervals

 

 

Large Sample Confidence Intervals

We construct confidence interval for the mean to give an indication of where the mean lies.  

 

Example:  

Suppose that we check for clarity in 50 locations in Lake Tahoe and discover that the average depth of clarity of the lake is 14 feet with a standard deviation of 2 feet.  What can we conclude about the average clarity of the lake?  

 

Solution

We can use x to provide a point estimate for m and s to provide a point estimate for s.  How accurate is x as a point estimate?  We construct a 95% confidence interval for m as follows.  We draw the picture and realize that we need to use the table to find the z-score associated to the probability of .025 (there is .025 to the left and .025 to the right).  

We arrive at z = -1.96.  Now we solve for x:

                          x - 14             x - 14
        -1.96  =                    =                                                   
                          2/              0.28                                                

Hence

        x - 14 = -.55

We say that +-.55 is the margin of error.

We have that a 95% confidence interval for the mean clarity is

        (13.45,14.55)

In other words there is a 95% chance that the mean clarity is between 13.45 and 14.55.

In general if z* is the z value associated with x% then an x% confidence interval for the mean is

       

 

  1. Calculating n

     

    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 s knowing s = 10.  How many people should we ask?

     

    Solution

    Solving for n in

            Margin of Error  =  B  =  z*s/

    we have

            B =  z*s

            = z*s/B

     n = [z*s/B]2  

     

    We use the formula:  

            n = (1.96(10)/0.5)2 = 1,536

    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 

            B = 3,       z* = 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:

            n = (1.65(13)/3)2 = 51.1

    Hence the dealer should survey about 52 people.

Handout on finding the sample size for a confidence interval for a population mean

Back to the Estimation Home Page

Back to the Elementary Statistics (Math 201) Home Page

Back to the Math Department Home Page

e-mail Questions and Suggestions