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COMPARING TWO POPULATIONS. IDEA: Compare two groups/populations based on samples from each of them. Examples. Compare average height of men and women. Draw sample of men heights: x1, x2, …, xm and a sample of women heights: y1, y2, …, yn. Test Ho:

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Comparing two populations
COMPARING TWO POPULATIONS

IDEA: Compare two groups/populations based on samples from each of them.

Examples.

  • Compare average height of men and women. Draw sample of men heights: x1, x2, …, xm and a sample of women heights: y1, y2, …, yn. Test Ho:

    Ho: μx = μy vs Ha: μx ≠ μy or Ha: μx > μy

  • Compare proportions of Democrats in two cities,

  • Compare weights of people before and after a diet, etc.

    General considerations for the samples: Dependent or independent samples.

    Example. Comparing weights of people before and after a diet we have dependent(same people) samples of weights. Comparing weights of people in two cities we have independent samples. Analysis methods will differ for dependent and independent samples.


Paired t test dependent samples
PAIRED t-TEST: dependent samples

Observations come as matched pairs (X,Y).

X and Y are NOT independent, X and Y are dependent.

Examples.

  • X is score on a test before studying hard; Y is score on the test after studying hard for the same student;

  • X is score on a test or in sports before training program, Y score after training program;

  • X is weight before weight loss program, Y is weight after the program;

  • X and Y are heights of twins or siblings.


Paired t test hypotheses
PAIRED t-TEST: HYPOTHESES

Hypotheses of interest: does training make a difference?

μx = score before training; μy = score after training.

Ho: μx = μy vs Ha: μx < μy

(no difference) (score after training is higher)

Data are pairs of observations: (x1, y1), (x2, y2), …, (xn, yn).

Typically, we work with differences: d=X-Y, and phrase hypotheses in terms of differences: μd = true mean difference.

In terms of differences:

Hypotheses

e.g. Ho: μd = 0 vs Ha: μd < 0

Data: d1, d2, …, dn.


Paired t test test procedure
PAIRED t-TEST: TEST PROCEDURE

To test Ho, we do one sample t-test. Need sample mean and standard deviation of d’s:

Compute the test statistic:

Under Ho the test statistic has t(n-1) distribution.

Make decision in exactly the same way as for the one sample t-test.

A (1-α)100% CI for d:


Paired t test an example
PAIRED t-TEST: an example

The amount of lactic acid in the blood was examined for 10 men, before and after a strenuous exercise, with the results in the following table.

(a) Test if exercise changes the level of lactic acid in blood. Use significance level α=0.01.

(b) Find a 95% CI for the mean change in the blood lactose level.


Paired t test lactic acid example contd
PAIRED t-TEST: lactic acid example contd.

Solution. Take d=“After level” – “before level” of lactic acid.

Data for d: 18, 4, 17, 22, 23, 17, 5, 10, 7, 1. Sample stats:

STEP1.Ho: μd = 0 vs Ha: μd ≠ 0

STEP 2. Test statistic:

STEP 3. Critical value? df=n-1=9, tα/2 =t0.005=3.69.

STEP 4. DECISION: t = 4.93 > 3.69 = t0.005 , so reject Ho.

STEP 5. Exercise changes lactic acid level.


Example contd
Example contd.

(b) Find a 95% CI for the mean change in the blood lactose level.

It is the familiar formula for the 95% CI for the mean, this time mean difference μd. Need percentile from the t distribution with n-1 degrees of freedom.

n=10, n-1=9, α=0.05, so tα/2 =t0.025=2.262, so the 95% CI for μd is:



Lactic acid example in minitab data set lactic acid mpj1
Lactic acid example in MINITAB: data set lactic-acid.MPJ

Paired T-Test and Confidence Interval

Paired T for before - after

N Mean StDev SE Mean

before 10 15.50 2.37 0.75

after 10 27.90 8.17 2.58

Difference 10 -12.40 7.95 2.51

95% CI for mean difference: (-18.08, -6.72)

T-Test of mean difference = 0 (vs not = 0): T-Value = -4.93 P-Value = 0.001

Ho

Ha

Conclusion: Reject Ho, lactic acid level changes after exercise.

Note: CI for “Before –after”


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