1

Some terminology

Population - the set of all cases about which we have some interest.

Sample - the cases we have selected from the population (randomly)

Parameter - a characteristic of the population. E.g., average age

Statistic - a characteristic of the sample

2

Population and Sample

Population Parameter - unknown. For example mean age of population

Sample statistic - known because we calculate it. For example the mean age of the cases in the sample.

Is the sample mean a good estimate of the population mean?

3

Sample represent population?

Recall example of 3 red and 7 black marbles

We took sample of size 5 (n = 5), with replacement

Our sample could contain 0, 1, 2, 3, 4, or 5 red marbles

4

Probability of r red marbles

00.05

0.10.15

0.20.25

0.30.35

0.4

0 1 2 3 4 5

Number of Red Marbles

Pro

ba

bilit

y

Sampling Distribution

5

What if

Sample had 0 red marbles? Surprised? No! This could happen 17% of the time with random samples from a population with 3 red and 7 black marbles.

Sample had 1 red marble? Surprised? No! This could happen 36% of the time with random samples….

6

What if

Sample had 2 red marbles? Surprised? No! This could happen 31% of the time with random samples from a population with 3 red and 7 black marbles.

Sample had 3 red marble? Surprised? Not really. This could happen 13% of the time with random samples….

7

What if

Sample had 4 red marbles? Surprised? I think so. This could happen only 3% of the time with random samples….

Sample had 5 red marble? Surprised? Very! This could happen only .2% of the time with random samples from a population with 3 red and 7 black marbles.

8

Note our logicIf the population has certain characteristics

then our sample will probably include certain outcomes and probably not include other outcomes

If our sample has outcomes that are unlikely to come from that population, it probably did not come from that population

9

Our hypothesis is

The population has certain characteristics.If our sample is unlikely to have come from that population, reject the hypothesis.If our sample could easily have come from that population, fail to reject the hypothesis

10

The question then is

How can we tell how likely we are to get a particular result (sample with certain characteristics) from a particular population (with its characteristics)?What are the rules for doing this?We have already done it once with the red and black marbles. Let’s look at that again

11

Population characteristics

3 red marbles

7 black marbles

random sample with replacement

Hence p(r) = .3, p(b) = .7

12

Sample characteristics

Random sample with replacement of n=5

Count number red marbles

Hence outcomes could be 0, 1, 2, 3, 4, or 5

Find probabilities of every possible outcome

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Find probabilities

All black = .7 * .7 * .7 * .7 * .7 = .17

One red = 5 * .3 * .7 * .7 * .7 * .7 = .36

Two red = 10 * .3 * .3 * .7 * .7 *.7 = .31

Three red = 10 * .3 * .3 * .3 * .7 * .7 = .13

Four red = 5 * .3 * .3 * .3 * .3 * .7 = .03

Five red = .3 * .3 * .3 * .3 * .3 = .002

Total = .17+.36+.31+.13+.03+.002= 1.00

14

Probability of r red marbles

00.05

0.10.15

0.20.25

0.30.35

0.4

0 1 2 3 4 5

Number of Red Marbles

Pro

ba

bili

ty

Unlikely

15

Try a new one

Population has 10 persons with ages 20, 20, 30, 30, 40, 40, 50, 50, 60, and 60 years (uniform population)

Mean age of population is 40 years

Random sample of n=3 (with replacement)

What outcomes might we expect?

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All possible sample outcomes

20, 20, 20: m = 20 : p= 1/5*1/5*1/5

20, 20, 30: m = 23.33: p= 3*1/5*1/5*1/5

20, 20, 40: m = 26.67: p= 3*1/5*1/5*1/5

20, 30, 30: m = 26.67: p= 3*1/5*1/5*1/5

Etc.

60, 60, 60: m = 60: p = 1/5*1/5*1/5

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Group according to means

20, 23.33, 26.67, 30, 33.33, 36.67, … , 56.67, 60Examine probability of each possible meanA lot of work, and luckily we do not have to do that each time -- indeed, we could not, because we do not know all the values in the population

18

We assume population shape

The age distribution was a rectangular probability distribution (all ages were equally likely).

Let’s see what happens when we draw random samples of size n=3 and calculate the mean

19

Output of program samp sam.mean obs1 obs2 ob3 mean 1 46.66667 40 50 50 46.6 2 30 30 40 20 38.3 3 46.66667 60 40 40 41.1 4 36.66667 60 20 30 40 5 46.66667 30 60 50 41.3 6 30 30 20 40 39.4 7 50 40 50 60 40.9 8 26.66667 30 30 20 39.1 9 43.33333 40 60 30 39.6 10 36.66667 40 50 20 39.3 11 40 40 20 60 39.3 12 43.33333 20 60 50 39.7 13 36.66667 20 50 40 39.4

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ResultsFraction

mean23.3333 56.6667

0

.1666

Sampling Distribution

21

Mean of meansmeanmean

sample1 5000

38.3333

46.6667