Estimating Differences

Difference Between Means

I surveyed 50 people from a poor area of town and 70 people from an affluent area of town about their feelings towards minorities.  I counted the number of negative comments made.  I was interested in comparing their attitudes.   The average number of negative comments in the poor area was 14 and in the affluent area was 12.  The standard deviations were 5 and 4 respectively.  Let's determine a 95% confidence for the difference in mean negative comments.  First, we need some formulas.

Theorem The distribution of the difference of means x1 - x2 has mean           m1 - m2  and standard deviation

For our investigation, we use s₁ and s₂ as point estimates for s₁ and s₂.  We have

        x₁  =  14        x₂  =  12        s₁  =  5        s₂  =  4        n₁  =  50        n₂  =  70

Now calculate

        x₁  -  x₂  =  14 - 12  =  2

The margin of error is

        E  =  z_cs  =  (1.96)(0.85)  =  1.7

The confidence interval is

        2  1.7  

or

        [0.3, 3.7]

We can conclude that the mean difference between the number of racial slurs that poor and wealthy people make is between 0.3 and 3.7.

Note:  To calculate the degrees of freedom, we can take the smaller of the two numbers n₁ - 1 and n₂ - 1.  So in the prior example, a better estimate would use 49 degrees of freedom.  The t-table gives a value of 2.014 for the t_.95 value and the margin of error is

                E  =  t_cs  =  (2.014)(0.85)  =  1.7119

which still rounds to 1.7.  This is an example that demonstrates that using the t-table and z-table for large samples results in practically the same results.

Small Samples With Pooled Pooled Standard Deviations (Optional)

When the sample size is small, we can still run the statistics provided the distributions are approximately normal.  If in addition we know that the two standard deviations are approximately equal, then we can pool the data together to produce a pooled standard deviation.  We have the following theorem.

You've gotta love the beautiful formula!

Note

After finding the pooled estimate we have that a confidence interval is given by

Example

What is the difference between commuting patterns for students and professors.  11 students and 14 professors took part in a study to find mean commuting distances.  The mean number of miles traveled by students was 5.6 and the standard deviation was 2.8.  The mean number of miles traveled by professors was 14.3 and the standard deviation was 9.1.  Construct a 95% confidence interval for the difference between the means.  What assumption have we made?

Solution

We have

        x₁  =  5.6        x₂  =  14.3        s₁  =  2.8        s₂  =  9.1        n₁  =  11        n₂  =  14

The pooled standard deviation is

The point estimate for the mean is

        14.3 - 5.6  =  8.7

and

Use the t-table to find t_c for a 95% confidence interval with 23 degrees of freedom and find

         t_c  =  2.07

        8.7 (2.07)(7.09)(.403)  =  8.7 5.9

The range of values is  [2.8, 14.6]

The difference in average miles driven by students and professors is between 2.8 and 14.6.  We have assumed that the standard deviations are approximately equal and the two distributions are approximately normal.

Difference Between Proportions

So far, we have discussed the difference between two means (both large and small samples).  Our next task is to estimate the difference between two proportions.  We have the following theorem

And a confidence interval for the difference of proportions is

Confidence Interval for the difference of Proportions

Note:  in order for this to be valid, we need all four of the quantities

        p₁n₁         p₂n₂         q₁n₁         q₂n₂ 

to be greater than 5.

Example

300 men and 400 women we asked how they felt about taxing Internet sales.  75 of the men and 90 of the women agreed with having a tax.  Find a confidence interval for the difference in proportions.

Solution

We have

        p₁  =  75/300  =  .25        q₁  =  .75        n₁  =  300

        p₂  =  90/400  =  .225        q₂  =  .775        n₂  =  400

We can calculate

We can conclude that the difference in opinions is between -8.5% and 3.5%.

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