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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**

**=============================================**

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

A The ten sample differences (# mg. in name brand - #mg. in generic brand) are as follows:
Here is a dotplot displaying the differences.
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
The calculated t statistic is
At the 5% level of significance, the t-distribution table
indicates that H 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,
H |

**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 H
_{0}should be rejected. - One could note that there are 8 positive differences, 1
negative difference, and 1 zero difference. If H
_{0}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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