Please turn off cell phones, pagers, etc.The lecture will begin shortly.

There will be a quiz at the end of today’s lecture.

Lecture 34

Today’s lecture will cover material that is not in the textbook.

2. Difference in means for paired data

3. Regression to the mean

1. Review CI for a difference between means

1. Review CI for difference in means

Last time, we learned how to construct a CI and test for the difference between two population means.

Population #1:

μ

σ

1

1

Population #2:

μ

σ

2

2

These methods assume that we have independent samples from the two populations.

Compute the difference between the sample means:

diff = mean from sample #1 – mean from sample #2

Then compute the standard error of the difference:

SE diff = square root of

(SE of first mean) + (SE of second mean)

2 2 [ ]

How to do it

The 95% confidence interval is diff plus or minus 2 × SE of diff

If this confidence interval doesn’t cover zero, we can reject the null hypothesis that the two population meansare equal.

Example

An experiment was conducted to assess the manual dexterity of male and female factory workers. On average, men took longer to assemble a product than women, and their standard deviation was also higher.

Average

Standard deviation

Sample size

30.3

12.5

32

25.3

10.0

40

men women

Can we conclude beyond a reasonable doubt that the population average time is longer for men than for women?

Solution

Average

Standard deviation

Sample size

30.3

12.5

32

25.3

10.0

40

men women

diff = 30.3 – 25.3 = 5.0

SE for first mean = 12.5 / sqrt( 32 ) = 2.21SE for second mean = 10.0 / sqrt( 40 ) =

1.58SE diff = sqrt( 2.21 + 1.58 ) = 2.72

2 2

5.0 – 2 × 2.72 = – 0.44 5.0 + 2 × 2.72 = 10.44

The confidence interval (– 0.44, 10.44) covers zero.So we do not have enough evidence to conclude that thepopulation means for men and women are different.

2. Difference in means for paired data

In some studies, the same subjects are being measured twice. Or, each subject in sample #1 may have a pairwise relationship with a subject in sample #2 (spouses, siblings, twins, etc.)

The method we just learned assumes that the two samples are independent, i.e. that there are no strong connections between the subjects in one sample and the subjects in the other sample.

In that case, we would say that we have paired data.

We will now learn how to form a confidence interval for the difference between the means from paired samples.

Step 1: Compute the change scores

5.1

first second Measurement:

4.3

6.7

12.5

1

2

3 …

n

…

4.2

6.6

2.7

9.5 …

change

0.9

–2.3

4.0

3.0

…

For each pair, compute the change score.

The change score is simply the difference between the first and second measurements.

Step 2: Find the average and SD of the change scores

change

0.9

–2.3

4.0

3.0

…

Find the average of these numbers in the usual way (add them up, divide by n).

Find the SD of these numbers in the usual way (use a computer, or use the three-column method from Lecture 14).

We now have the average change score and the SD of the change scores.

Step 3: Find a 95% CI for the population average change

The SE for the average change is

If this 95% CI does not cover zero, then we can reject the null hypothesis that there is no average change in the population.

SE = SD of the change scores / sqrt( n )

The 95% CI is then

average change plus or minus 2 × SE

ExampleFor six months, forty hypertensive subjects were given an herbal supplement that is supposed to help reduce blood pressure. The average change in systolic BP (after minus before) was –2.2, with a standard deviation of 3.6. Is this convincing evidence that the supplement works? SolutionWe have already been given the average change score and the SD of the change scores.

The SE of the average change is SE = 3.6 / sqrt( 40 ) = 0.57, and two SEs is 2 × 0.57 = 1.14.

The 95% CI for the population average change is–2.2 – 1.14 = –3.34 to –2.2 + 1.14 = –1.06.

This interval lies entirely below zero. This provides convincing evidence that, on average, subjects similar to those in the study who take the supplement will experience a drop in BP.

Discussion

In the previous example, we concluded that the average population change was not zero.

Does this prove that the herbal supplement works?

It does prove that, if the supplement were given to a large population of males similar to those in the study, on average they would experience a drop in BP. But this does not prove that the supplement is any more effective than taking a placebo or doing nothing!This study had no control group. So there is no firm basis to claim that the supplement is effective.

In fact, if these subjects were hypertensive to begin with, then it is highly likely that they would experience a drop in blood pressure even if they were given no treatment at all!

3. Regression to the mean

“Regression to the mean” is a phenomenon first noticed by Sir Francis Galton in the 1870’s.

Suppose a sample of units is measured twice, and the measurements are positively correlated.

Units that score above average on the first measurement also tend to score above average on the second measurement, but their average score is not as high as it was the first time.

Units that score below average on the first measurement also tend to score below average on the second measurement, but their average score is not as low as it was the first time.

Example

Suppose you take a sample of adult men and identify a group whose heights are above average. If you examine their sons, you will find that the sons of these men are also above average in height, but on average they are not as tall as their fathers.

Students in Stat 100 who scored very well on Exam 1 also tended to score well on Exam 2, but their average score on Exam 2 was lower than their average score on Exam 1.

Example

What’s happening?

Regression to the mean will occur in any test-retest situation.Whenever you identify units who score high on the first test, some of them will have high scores just by chance. If these same units are measured a second time, the laws of probability suggest that they are unlikely to do as well as they did the first time.

Regression to the mean does not indicate that, in the long run, everyone’s scores will converge to the population average. Random fluctuations are still present in every wave of measurement.

One more example

Suppose that you toss 100 pennies into the air and 50 of them land “heads”.

You collect those 50 pennies and lecture them: “You pennies are very naughty. Half of you were supposed to be tails, but none of you were. Make sure that you do a better job next time!”

You toss those 50 pennies into the air again, and you find that half of them fall “heads” and the other half fall “tails.”Your lecture cured them of their bad behavior!

Of course, your lecture had no effect whatsover. Those pennies were not naughty at all. They were just behaving as pennies always do. It was regression to the mean.

Today’s quiz

1. Write your name legibly.