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.
For our investigation, we use s_{1} and s_{2} as point estimates for s_{1} and s_{2}. We have x_{1} = 14 x_{2} = 12 s_{1} = 5 s_{2} = 4 n_{1} = 50 n_{2} = 70 Now calculate x_{1}  x_{2} = 14  12 = 2
The margin of error is E = z_{c}s = (1.96)(0.85) = 1.7 The confidence interval is 2 1.7 or [0.3, 3.7]
Note: To calculate the degrees of freedom, we can take the smaller of the two numbers n_{1}  1 and n_{2}  1. So in the prior example, a better estimate would use 49 degrees of freedom. The ttable gives a value of 2.014 for the t_{.95} value and the margin of error is E = t_{c}s = (2.014)(0.85) = 1.7119 which still rounds to 1.7. This is an example that demonstrates that using the ttable and ztable 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
x_{1} = 5.6 x_{2} = 14.3 s_{1} = 2.8 s_{2} = 9.1 n_{1} = 11 n_{2} = 14 The pooled standard deviation is
The point estimate for the mean is 14.3  5.6 = 8.7 and
Use the ttable 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
Note: in order for this to be valid, we need all four of the quantities p_{1}n_{1} p_{2}n_{2} q_{1}n_{1} q_{2}n_{2} 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 p_{1} = 75/300 = .25 q_{1} = .75 n_{1} = 300 p_{2} = 90/400 = .225 q_{2} = .775 n_{2} = 400 We can calculate
We can conclude that the difference in opinions is between 8.5% and 3.5%.
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