Example of MATCHED PAIRS t PROCEDURE

Sanderson M. Smith

Example of MATCHED PAIRS t PROCEDURE
(Problem #5 on 2001 AP Statistics Examination)

This is an analysis of Problem #5 in Section II on the 2001 Advanced Placement Statistics Examination.

The College Board does not allow a reproduction of the problem statement here. If you do not have before you a statement of the actual problem, you can get it from the College Board site

http://www.collegeboard.org/ap/statistics/frq01/index.html

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Here is a solution to Problem #5 on the 2001 AP STATISTICS EXAMINATION.

A paired t test of differences is appropriate here.

The ten sample differences (# mg. in name brand - #mg. in generic brand) are as follows:

-1
4
5
13
0
7
5
8
9
16

Here is a dotplot displaying the differences.

*

*

*

*

*

*

*

*

*

*

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

Assumptions check: The differences represent a random sample from a population of differences (premise states pharmacies were randomly selected), and the distribution of differences is approximately normal. A modified box plot (TI-83) shows no outliers.

Using TI-83, we find mean of sample differences = 6.6, with s = 5.27. If m_d =the population mean of sample differences, then the null and alternate hypotheses are

H₀: m_d = 0.

H_a: m_d is not 0. (Problem statement requires a two-sided test).

The calculated t statistic is

t = (6.6 - 0)/(5.27/÷(10)) = 3.96, with 10-1 = 9 degrees of freedom.

At the 5% level of significance, the t-distribution table indicates that H₀ would be rejected if t > 2.262 or if t < -2.262. Our calculated t statistic of t = 3.96 is in the critical region, so we reject H₀ at the 5% significance level. The consumer group should report that the difference (in milligrams) between the amount of active ingredient in the name brand and the amount in the generic brand is significant.

We can further note that the p-value (calculated from TI-83) for t = 3.96 with 9 degrees of freedom is 2*tcdf(3.96,1E99,9) = 0.0033 = 0.33%. In other words, H₀ would be rejected at the 1% level of significance, and even at the 1/2% level of significance.

Additional notes relating to this problem:

Two alternate, potentially correct approaches, are shown below. Note that one would have to apply appropriate details and explanations to get complete credit for such solutions.

One could take a confidence interval approach. For instance, a 95% paired t confidence interval for the mean of the differences is 6.6 plus/minus 2.262( 5.27/÷(10). Doing the computation, this interval is (2.83, 10.37). Since this interval does not contain 0, there is evidence that H₀ should be rejected.
One could note that there are 8 positive differences, 1 negative difference, and 1 zero difference. If H₀ is true, then it is reasonable to assume that the probability of a positive difference is 0.5. Ignoring the zero difference and using the TI-83, the probability of having at least 8 positive differences is 1 - binomcdf(9,.5,7) = 0.01953125, or approximately 2%.

The following statistical tests would not be appropriate for this problem.

A z test: A sample size of 10 is too small. In general, the sample size should be at least 30 before a z test is appropriate.
A chi-square test: This test deals with counts (discrete data) and not measurements such as those in this problem that represent continuous data. While the amounts (mg.) in the chart are integers, a number such as 245 is rounded, and could represent any number that is at least 244.5 and less than 245.5.
A two-sample t test: This requires that the two samples involved are independently selected random samples. This condition is not met.
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