Matched Pairs t-procedures:

Subjects are matched according to characteristics that affect the response, and then one member is randomly assigned to treatment 1 and the other to treatment 2. Recall that twin studies provide a natural pairing. Before and after studies are examples of matched pairs designs, but they require careful interpretation because random assignment is not used.

Apply the one-sample t procedures to the differences

Confidence Intervals for Matched Pairs

n

Stx dnd*

1

Example #1Archaeologists use the chemical composition of clay found in pottery artifacts to determine whether different sites were populated by the same ancient people. They collected five random samples from each of two sites in Great Britain and measured the percentage of aluminum oxide in each. Based on these data, do you think the same people used these two kiln sites? Use a 95% confidence interval for the difference in aluminum oxide content of pottery made at the sites and assume the population distribution is approximately normal. Can you say there is no difference between the sites?

New Forrest

20.8 18 18 15.8 18.3

Ashley Trails

19.1 14.8 16.7 18.3 17.7

Difference 1.7 3.2 1.3 -2.5 .6

P: μn = New Forrest percentage of aluminum oxide

The true mean difference in aluminum oxide levels between the New Forrest and Ashley Trails.

μa = Ashley Trails percentage of aluminum oxide

μd = μn - μa = Difference in aluminum oxide levels

A: SRS:

Normality:

Independence:

N: Matched Pairs t-interval

Says randomly selected

Says population is approx normal

It is safe to assume that there are more than 50 samples available

I: df =

n

Stx dnd*

1

5 – 1 = 4

I: df = 20 – 1 = 19

n

Stx dnd*

1

5

105469069.2776.286.

613866034.286.

4743.3 ,754.1

C: I am 95% confident the true mean difference in aluminum oxide levels between the New Forrest and Ashley Trails is between –1.754 and 3.4743.

Can you say there is no difference between the sites?

Yes, zero is in the confidence interval, so it is safe to say there is no difference.

Example #2The National Endowment for the Humanities sponsors summer institutes to improve the skills of high school language teachers. One institute hosted 20 Spanish teachers for four weeks. At the beginning of the period, the teachers took the Modern Language Association’s listening test of understanding of spoken Spanish. After four weeks of immersion in Spanish in and out of class, they took the listening test again. (The actual spoken Spanish in the two tests was different, so that simply taking the first test should not improve the score on the second test.) Below is the pretest and posttest scores. Give a 90% confidence interval for the mean increase in listening score due to attending the summer institute. Can you say the program was successful?

Subject Pretest Posttest Subject Pretest Posttest

1 30 29 11 30 32

2 28 30 12 29 28

3 31 32 13 31 34

4 26 30 14 29 32

5 20 16 15 34 32

6 30 25 16 20 27

7 34 31 17 26 28

8 15 18 18 25 29

9 28 33 19 31 32

10 20 25 20 29 32

P: μB = Pretest score

The true mean difference in test scores between the Pretest and Posttest

μd = μB - μA = Difference in test scores

μA = Posttest score

A: SRS:

Normality: When you are given the data you can graph it – use Normal Probability Plot

We must assume the 20 teachers are randomly selected

A: SRS:

Normality:

Independence:

N: Matched Pairs t-interval

We must assume the 20 teachers are randomly selected

15<n<30 and distribution is approximately normal, so safe to assume

It is safe to assume that there are more than 200 Spanish teachers

I: df =

n

Stx dnd*

1

20 – 1 = 19

I: df = 20 – 1 = 19

n

Stx dnd*

1

3.20321.45 1.729

20

1.45 1.2384

2.689, 0.2115

C: I am 90% confident the true mean difference in test scores between the Pretest and Posttestis between –2.689 and –0.2115.

Can you say the program was successful?

Yes, zero is not in the confidence interval, so the pretest score is lower than the posttest score.