• population
• samples
• the distribution of sample means

So far we've pretty much focused on descriptive statistics, which are ways of describing distributions (e.g., means and standard deviations).

The goal of inferential statistics is to make claims about population parameters based on sample statistics.

Typically we can't measure the entire population of individuals that we're interested in. So instead we select a sub-set of individuals, which we call a sample. Then we measure our sample, and use those measurements to make estimates of population values (which correspond to those measurements, e.g., means and standard deviations). Our best estimate for the mean of the population will be the mean of our sample.

It sounds simple and straight forward, but consider the following:

Suppose that you take 3 different samples from the same population. They are going to be different from one another. They will have different shapes, different means, and different variability. So how do you figure out what the best estimate of the population mean is? How many possible samples can we take? Infinite (if we are sampling with replacement)

Luckily for us, the huge set of possible samples forms a simple, orderly, and predictable pattern (a sampling distribution). Because of this, we are able to base our predictions about sample characteristics on the distribution of sample means.

The distribution of sample means

The distribution of sample means is a hypothetical collection of sample means for all the possible random samples of a particular size (n) that can be obtained from a population. In other words, what we want to do is look at all of the possible samples (of a particular size, this part is important) and make predictions based on the properties of all of them.

I consider this distribution to be virtual, that is we don't actually compute all possible samples of sample size n, but instead generate it based on properties of the population.

A simple example

Consider the following population of scores: 2, 4, 6, 8

Because this population is so small we actually can know the mean (and variability): m = 5, but suppose that we didn't, and wanted to be able to make an estimate based on sampling

step 1: pick a sample size: for this example we'll pick samples of n = 2
- we'll talk more about sample size a little later, but typically the bigger your sample size, the more likely that your samples will be similar to one another (and to the population as a whole)

step 2: now consider all of the possible samples that you could get, and look at their distribution

__________________________________ scores sample mean sample first second () 1 2 2 2 2 2 4 3 3 2 6 4 4 2 8 5 5 4 2 3 6 4 4 4 7 4 6 5 8 4 8 6 9 6 2 4 10 6 4 5 11 6 6 6 12 6 8 7 13 8 2 5 14 8 4 6 15 8 6 7 16 8 8 8

f 2 1 3 2 4 3 5 4 6 3 7 2 8 1

Distribution of the sample means

Properties of the distribution of sample means

shape:
the shape of the distribution of sample means tends to be a Normal distribution. In fact, when n is large (around 30 or more), the distribution of sample means is almost perfectly Normal.

Does this make sense? It should. If you select a bunch of samples from the same population, the most of the means should "pile up" near the population mean m (if they don't then you must have some kind of bias in your sample)

mean:

the average of all of the sample means will equal the mean of the population. The average of all of the sample means is called the expected value of . It is "expected" because it should be a value near the population mean m . (note: often symbolized as: m , but your book will usually just use m )

Look at our example, the expected value of (the mean of the sample means) is:

   
        2 + 3 + 4 + 5 + 3 + 4 + 5 + 6 + 4 + 5 + 6 + 7 + 5 + 6 + 7 + 
8 = 80 = 5.0
                                        16                              16

variability: the standard deviation of the distribution of sample means is called the standard error of
standard error of = = standard distance between and m .
in other words, this statistic describes the standard (typical/average) distance from the mean. In this case it is the distance between the sample mean and the population mean m .

the major purpose/use of the standard error of is that it tells us how well the sample mean estimates the population mean.
In other words, how big the sampling error is.

Computing the standard error

the long way
compute the standard deviation of the distribution of sample means

___________________________________________________
            scores     sample mean
sample	first   second    ()	(- m)	 (- m)2
1         2       2         2	  -3		9
2         2       4         3	  -2		4
3         2       6         4	  -1		1
4         2       8         5	   0		0
5         4       2         3	  -2		4
6         4       4         4	  -1		1
7         4       6         5	   0		0
8         4       8         6	   1		1
9         6       2         4	  -1		1
10        6       4         5	   0		0
11        6       6         6	   1		1
12        6       8         7	   2		4
13        8       2         5	   0		0
14        8       4         6	   1		1
15        8       6         7	   2		4
16	  8       8         8	   3		9 
					SS =	40

std dev = sqrt(SS/N) = sqrt(40/16) = sqrt(2.5) = 1.58

the short way

• the numerical value of the standard error is determined by two characteristics: the variability of the population & the size of the sample
  1) the variability of the population - the bigger the variability of the population, the more variability you'll have in the sample means.

  large s big differences from the pop mean

  small s small differerences from the pop mean

  2) the size of the sample - the larger your sample size (n), the more accurately the sample represents the population. This is known as the law of large numbers .

  think of it this way:
  suppose that you have a population of 1,000 students. You want to know the average SAT score of the entire population.

  	- If I randomly selected 1 student, how accurately will that student's score predict the population's score?
  	- Suppose that I take 5 students. Are things more accurate?
  	- what about 100 students?

  these two characteristics are combined in the formula for standard error.

  standard error of = =

  Returning to our example. The standard deviation of our original population is s = 2.24.

  So: 2.24/sqrt(2) = 1.58

  ---

  In our example we've simplified things greatly. We have a really small population, and we took a pretty small sample. Most situations, however, will be much more complex. Lucky for us there are some properties of means of samples that will help us out.

  All of these properties (shape, mean, variability) are covered in the Central Limit Theorem

  Central Limit Theorem: For any population with mean m and standard deviation s, the distribution of sample means for sample size n will approach a normal distriution with a mean of m and a standard deviation of as n approaches infinity.

  So, when n is large (at or above 30): ~ N (m , )
  

  (reminder: n = the number of scores in the sample)

Normal distribution is a commonly found distribution that is symmetrical and unimodal. It is defined by the following equation:

                                Y = 

A few things to note about Normal Distributions.

• Not all unimodal, symmetrical curves are normal, but a lot are
• For this class, we won't worry about how close a distribution is to normal, in fact for most of the course we'll assume that the distribution is normal
• A smooth curve like that above is refered to as a density curve (rather than a frequency curve)
• The area under (any density) curve must sum to 1. Why? remember that the area under the curve refers to the probabilities (or proportions) and the total probability must equal 1.
• The normal distriution is often transformed into z-scores.
• For a normal distribution:

  This is sometimes referred to as the 68-95-99.7 rule. In the normal distribution with mean m and a standard deviation s:
  34.13% of the scores will fall between m and 1 stdev. 13.59% of the scores will fall between 1stdev & 2stdev. 2.28% of the scores will fall between the 2stdev & 3stdev.	68% of the observations fall within s of the mean m. 95% of the observations fall within 2s of the mean m. 99.7% of the ovservations fall within 3s of the mean m.

• An important tool that we'll use is the unit normal table. You'll find it in the back of your book (Table B) (note: the on-line table and the table in your book are aranged slightly differently, but the information is the same in both). In this table are a bunch of z-scores and proportions for the Standard Normal Distribution (which is the z-score standarized Normal distribution). In other words this table allows you to figure out the area under the curve (and thus the probability of sampling) at nearly every position on the curve (defined in z-scores).

  Using the unit normal table.

  first column - the Z-score in question
  the rest of the columns - the p(Z < z) - the proportion of the distribution to the left of the z-score. The heading of the columns are the second decimal digit.
  

  z .00 .01 -3.4 -3.3 : : 0.0 : : 1.0 : : 3.3 3.4 0.0003 0.0005 : : 0.5000 : : 0.8413 : : 0.9995 0.9997 0.0003 0.0005 : : 0.5040 : : 0.8438 : : 0.9995 0.9997

  So by using the table, we can an ask about different areas under the curve. We can also go in both directions. That is, from the table of z-scores to probabilities and/or from probabilities to z-scores.

  Notice that z = 1.0 = .5000 + .3413 = the median + the 34.13% that we mentioned before So by using the table, we can an ask about different areas under the curve. And similar to last chapter, we can go in both directions. That is, from the table of z-scores to probabilities and/or from probabilities to z-scores.

  Note: Don't underestimate the value of drawing a picture of the distribution and trying to just "eyeball" the answer (in addition to doing the math). It just may save you from making a mistake.

Let's return briefly to our simplified example.

Now we can answer questions like: What is the probability of getting a sample with a mean greater than 7? p( > 7) = ?



look at our distribution of sample means, we find that 1 out of 16 have a mean greater than 7. So that's our answer: 1/16 = .0625 = 6.25%

Using the Unit Normal Table to answer questions of probability

Here is the "best" way to find a probability from the table:

step 1: sktech the distribution, showing the mean & standard deviation
step 2: sketch the score in question, being sure to place it on the correct side of the mean & roughly the correct distance from the mean
step 3: read the problem again to see if you need the probability of getting a score > or <. Shade this area on your sketch.
step 4: translate the X score into a Z-score
step 5: Use the correct column (and sign) to find the probability in the unit normal table.

Examples:

What is the probability of having an IQ of 130 or above? p(X > 130)? for IQ scores m = 100, s =15 z = (130 - 100)/15 = 2.0 --look at the table--> need Column C p = 0.0228
What is the probability of having an IQ of 85 or less? p(X < 70)? for IQ scores m = 100, s =15, z = (70 - 100)/15 = -1.0 --look at the table--> need Column C p = 0.1587

step 1: Sketch the normal distribution
step 2: shade the region corresponding to the required probability
step 3: locate the probability in the correct column of the table
step 4: label the edge of the shaded region with the z-score from the step above
step 5: compute the corresponding raw score (X).

** keep in mind that the percentile rank is equal to the probability of being at or below a given score. Thus, percentile ranks less than 50% refer to the lower tail.
Example:

What IQ score do you need to have to be in the top 5% of the population? The upper-tail is needed. p = 0.05 ---- look at the table ---> z = 1.65 so X = (1.65)(15) + 100 = 124.75

Sometimes we need to find the probability that X will fall between two scores rather than simply above a score or below a score.

step 1: Sketch the curve & shade the region of interest
step 2: Translate both scores to Z-scores
step 3: Look up the probabilities of scoring < or > each of the two z-scores
step 4: Add (or subtract) the probabilities accordingly

Example:

What is the prob. of scoring between 300 and 650 on the SAT? recall: m = 500, s =100 p(z < (650 - 500) = p(z < 1.5) = 0.9332 100 p(z < (300 - 500) = p(z < -2.0) = 0.0228 100 the .9332 from 650 includes the lower tail, so we determine the proportion in the lower tail, and subtract that p(300 < z < 650) = .9332 - .0228 =.9104

Example:

What is the prob. of scoring lower than 300 or higher than 650 on the SAT? recall: m = 500, s =100 p(z > (650 - 500) = p(z > 1.5) = 0.0668 100 p(z < (300 - 500) = p(z < -2.0) = 0.0228 100 the two numbers both reflect the proportions in the tails, so we just need to add them together p(300 < z < 650) = .0668 + .0228 =.0896

Examples:

What is your percentile rank if you have an IQ of 130? for IQ scores m = 100, s =15 z = (130 - 100)/15 = 2.0 --look at the table--> need Column B p = 0.9772 --> percntile rank 97.72
What is the interquartile range for the SAT? recall: m = 500, s =100 --look at the table --> find 25% & 75% 0.25 = a Z-score of -0.67 0.50 = a Z-score of +0.67 X = Zs + m = (-.67)(100) + 500 = 433 = (+.67)(100) + 500 = 567 IQR = 567 - 433 = 134

Note there is a short-cut for figuring out the IQR. Since the range is always + .67s, then you can compute the IQR as being (2)(.67)(m)

example: for SAT: (2)(.67)(100) = 134

Now let's bring Samples back into the picture.

Example:

Consider the following situation. An instructor is interested in the IQ of her students. She has 9 students in her class and thinks that they are, on the average, really smart. What is the probability that the group of students has a mean greater than or equal to 112?

recall for standardized IQ test: m = 100, s = 15

First we need to get the distribution of the samples (note: we'll assume a normal distribution even though n is less than 30.) ~ N (m, ) = N(100, 5)

Now we need to figure out the z-score that corresponds to this sample mean: the z-score pretty much looks like what we've used before: Z =


so for our example:

P( > 112) = P(Z > (112 - 100)/ 5 ) = P(Z > 2.4) = 0.0082

Does this answer make sense?

	- at first it looks wrong - it seems like 112 should be less than a z = 1, because 115 is where z should equal 1
	- however, we must remember that this isn't the correct distribution to be looking at, we need to look at the distribution of sample means. -we know that the distribution of sample means has a standard error = 5 and a mean = 100. - So 112 should have a z >2

Example:


How high a mean would a group of 25 have to have on IQ to be in the top 10% of the IQ distribution for groups of this size?
First we need to get the distribution of the samples (note: we'll assume a normal distribution even though n is less than 30.)
~ N (m, ) = N(100, 3)

Now we need to figure out the mean that corresponds to this range:


the formula looks pretty much looks like what we've used before:
= Z * + m = (= Z)() + m

so for our example:

step 1: look at unit normal table for 90%

step 2: = 1.28 * + 100 = (1.28)(3) + 100 = 103.84

so, for a group of 25 people, they'd have to have a mean of just under 104 to be in the top 10%

Suppose that we asked the same question for a smaller sample, n = 16? How does the answer change?

step 1: look at unit normal table for 90%

step 2: = 1.28 * + 100 = (1.28)(3.75) + 100 = 104.80

so, for a group of 16 people, they'd have to have a mean of over 104 to be in the top 10%

What about other sample sizes?

n = 9


= 1.28 * + 100 = (1.28)(5) + 100 = 106.40

n = 4

= 1.28 * + 100 = (1.28)(7.5) + 100 = 109.60

n = 1

= 1.28 * + 100 = (1.28)(15) + 100 = 119.20

Note: the if n = 1, the standard error equals the population standard deviation