Statistics 221

1
Statistics 221

• Chapter 7
• Sampling and Sampling Distributions

2
Samples and Populations

• A population is the entire set of all elements of
  interest in a study. Examples all students in a
  university, all residents of a country, all
  registered voters, etc.
• A sample is a subset of that population.
• Numerical summaries of a population are called
  parameters numerical summaries of a sample are
  called statistics.

3
Statistical Inference

• In a typical study, a representative sample will
  be drawn from a population and statistics will be
  calculated.
• The statistics are used to draw inferences about
  the population as a whole.

4
Random samples

• If the sample was drawn using recognized random
  sampling techniques, it will be representative
  of the population, and therefore, the sample
  statistics should provide good estimates of the
  population parameters.

5
Making inferences about population means and
proportions

• It is common to make inferences about population
  means and proportions using sample statistics
• For example, we may take a sample of employees
  and ask them what their annual salary is and then
  compute an average. That sample average will be
  our best point estimate of what the populations
  average salary is.
• Or we may take a sample of employees as ask them
  whether they are in favor of flex-hours. The
  percentage or proportion of that sample who favor
  flex-hours is our best point estimate of the
  population proportion who favor flex-hours.

6
The Electronics Associates sampling problem

• The director of Personnel for Electronics
  Associates, Inc (EAI) has been assigned the task
  of developing a profile of the companys 2500
  managers.
• The characteristics of interest are
• Average salary
• The proportion who have completed the management
  training program.

7
Sampling techniques

• The most common method for gathering a sample is
  to use simple random sampling.
• The process of selecting a simple random sample
  depends on whether the population is finite or
  infinite.

8
Sampling from a finite population

• The population of managers at EAI is finite
  (2500).
• There are many possible samples of say size 30
  (n30) that could be drawn from a population of
  2500 (N2500).
• The sampling process should assure that each
  possible sample of size n drawn from population N
  has an equal chance of being selected.
• One technique would be to assign each manager a
  number and then use a random-number generator to
  generate n random numbers. If a managers number
  is generated, that manager is selected for the
  survey.

9
Sampling with and without replacement from a
finite population

• It is possible that a managers number is
  selected more than once. If we allow that to
  happen, we are sampling with replacement. If we
  eliminate that managers number so that it cant
  be selected again, we are sampling without
  replacement.
• For a finite population we generally follow a
  sample without replacement procedure by making
  sure that the same number is not selected more
  than once.

10
Sampling with and without replacement from an
infinite population

• When the population is infinite, the population
  size (N) is so large and the sample size (n) is
  so (relatively) small that the probability that a
  managers number will come up more than once is
  so small, so we assume that it wont happen and
  we just proceed as if we are sampling with
  replacement.

11
What were about to learn

• Were about to learn that if you took a large
  number of separate samples of a population, and
  you plotted all the samples means, you would
  have a normal distribution even if the
  populations distribution is NOT normal.
• The question we seek the answer to is still
  Whats the probability that x is x? but now x
  is a mean of sample not a single value.

12
Example

• A new establishment is open for only three days
  before it goes out of business. The sales volume
  for each of those three days was 1, 2, and 5.
• We want to statistically analyze sales volume.
  The population of interest consists of the values
  1, 2, and 5.

13
Calculate parameters

• The population is size 3.
• The mean ? of the population is (1 2 5)/3 or
  8/3 or 2.7.
• The std deviation ? of the population is 1.7.

• X (x-mean) (x-mean)2
• 1 - 1.7 2.89
• 2 - .7 .49
• 2.3 5.29
• 8.67

Sqrt(8.67 / 3) 1.7
14
Now let us take all possible samples of size 2

• We will list all possible samples of size 2 with
  replacement. Therefore, samples can consist of
  the same value twice (1-1, 2-2, 5-5).
• Why are we using with replacement? Because in
  most cases (but not in this case) the sample size
  is going to be less than 5 of the population,
  and when that happens, we use the with
  replacement formulas instead of the without
  replacement formulas because the with
  replacement formulas are simpler.

15
How many permutations of 2 are possible from
three values?

• 3 3 or 9
• Here are the 9 possible samples
• 1-1, 1-2, 1-5
• 2-1, 2-2, 2-5
• 5-1, 5-2, 5-5

16
Here are all possible samples of size 2 along
with their sample statistics
17
Important Point

• Notice that when we take the mean of the means,
  we get 2.7 that is also the population mean.
• When the group of samples includes all possible
  samples, then the mean of the means will always
  target the population mean.

18
Important Point
Frequency Distribution of the Sample Means
Although this distribution doesnt look so
normal, that is because the sample size is only
2. As the sample size increases, the probability
distribution of the samples means will approach
a normal distribution.
19
Example 2 (The unknown is the population
proportion / percentage)

• This time we consider a population that consists
  of all the US Senators -87 being male and 13
  being female.
• Now lets say we dont know that 13 are female
  because the population is too big to survey
  everyone.
• We are trying to determine the proportion of
  senators that are female. (In other words, the
  unknown is the proportion / percentage of the
  population that are female.)

20
1. We start out by obtaining several samples of
size 5

• We decide to take 100 samples of size 5 and
  record the percentage of female senators in each
  sample.
• We create a frequency distribution that shows the
  of females in each sample of 5.
• If we take the mean of all those samples
  percentages, we should get 13.

21
Results from just 100 samples of size 5

• of Females (x) Frequency (f) ( of samples)
• 0 26
• .1 41
• .2 24
• .3 7
• .4 1
• .5 1
• Mean (?(x f) .119 (not quite 13)

22
Why didnt we get 13 for the mean percentage of
women?

• Because we only took 100 samples of size 5, when
  there are a possible 1005 (10 billion) samples of
  size 5.
• So the mean of sample means didnt exactly mirror
  the population mean.

23
Here is the distribution of sample percentages
when the number of samples is 100
It resembles a normal curve but its not exactly
normal.
24
If our frequency distribution had included all
possible samples of size 5

• (1) the distribution of sample percentages would
  be (almost) normal and
• (2) the mean of sample means would have been the
  same as the population mean (13).

It would be (completely) normal only if n 30.
25
If we had taken all 10 billion possible samples
the distribution would be almost normal with a
mean of .13
26
Another important point

• We can see that when using a sample statistic to
  estimate a population parameter, some statistics
  are good in the sense that they target the
  population parameter and are therefore likely to
  yield good results. Such statistics are called
  unbiased estimators.
• Statistics that target population parameters
  mean, variance, proportion.
• Statistics that do not target population
  parameters median, range, standard deviation

27
Practice exercise (p. 256 - 6)

• Here are the numbers of sales per day that were
  made by Kim Ryan, a courteous telemarketer who
  worked four days before being fired 1, 11, 9, 3.
  Assume that samples of size 2 are randomly
  selected with replacement from this population of
  four values.
• A. List the 16 different possible samples and
  find the mean of each of them.
• B. Identify the probability of each sample, then
  describe the sampling distribution of sample
  means (Hint see Table 5-3).
• C. Find the mean of the sampling distribution
• D. Is the mean of the sampling distribution (from
  part c) equal to the mean of the population of
  the four listed values? Are those means always
  equal?

28
Practice exercise (p. 256 - 6)

• Open file dataSetsForProjectsCh5.xls
• Go to worksheet telemarketing
• Fill in the shaded cells with the appropriate
  values and answer the questions.

29
The Central Limit Theorem
30
The Central Limit Theorem conditions

• 1. Lets say that a random variable x has a
  distribution (which may or may not be normal)
  with mean µ and standard deviation ?.
• 2. Several samples all of the same size n are
  randomly selected from the population.
• 3. All the sample means are plotted on a
  probability distribution.

31
The Central Limit Theorem assertions

• 1. That probability distribution of sample means
  will, as the sample size increases, approach a
  normal distribution even if the population does
  not have a normal distribution!
• 2. Further, the mean of the sample means (?x )
  will be the same as the population mean µ.
• 3. The standard deviation of the sample means
  (?x) will be ? / ?n

32
One more point to add to the Central Limit theorem

• Recall 1 again the probability distribution of
  sample means will, as the sample size increases,
  approach a normal distribution even if the
  population does not have a normal distribution!
• The key word is approach. That is if the sample
  size (n) 30, then the distribution of sample
  means will be normal.
• If the sample size (n) is of the sample means will approach normal.
• BUT if the original population is already
  normally distributed, then the distribution of
  sample means will be normal for any sample size n
  (not just when n 30).

33
An example demonstrating the Central Limit
Theorem concept

• If we take the last 4 digits of the social
  security numbers of every US citizen, we have a
  population of values that form a uniform
  distribution.
• Recall that a uniform distribution means that
  every value from 0000 to 9999 is equally likely
  to occur.

34
A uniform distribution
35
Lets say we select a sample of 50 people

• And we take the last 4 digits of each of their
  social security numbers and we lump them
  together as a one big sample of 200 (4 50)
  digits.
• Then we calculate the mean of those 200 numbers
  to be 4.5
• Then we calculate the std deviation of those 200
  values to be 2.8.

36
Then we create a frequency distribution based on
that one sample of 200 digits
Distribution of 200 digits from Social Security
Numbers (Last 4 digits from 50 students)
Its not normal nor does it approximate the
uniform distribution of the population very
closely.
37
But now treat the sample data as 50 samples of 4
instead of 1 sample of 200
And calculate a mean for each sample of size
4then create a frequency distribution of those
sample means
38
And we have an (almost) normal distribution
Distribution of 50 Sample Means
Even though the population does not have a normal
distribution, the distribution of the sample
means is (almost) normal. And the std deviation
is ? / ? n.
39
Furthermore

• Had we used samples of size 30 or more (instead
  of 4) we could remove the word almost the
  sample means would have a frequency distribution
  that is fully-normal.
• As the sample size increases, the frequency
  distribution of the sample means approaches
  normal.

40
Applying the Central Limit Theorem

• In practice, we dont take several samples of
  size n we take one sample of size n and we treat
  the mean of that sample like its a single
  x-value - one of many possible sample means.
• Then we calculate a z-value for that x so that we
  can derive a p-value (the probability of getting
  that particular x (sample mean).
• Notice that we are still operating under the
  assumption that we know the population mean ? and
  population std deviation ? .
• We are still asking the question, Whats the
  probability of x being x? but now x is one of
  several potential sample means.

41
Calculating a z-value when x is a sample mean
not a single value
X as one possible single outcome
X as one possible sample mean
x - ?
x - ? x
z
z
?
? x

• Its actually the same formula because ?x ?
  (the mean of the population is the same as the
  mean of the sample means) but the standard
  deviations of the two distributions are
  different. The standard deviation for the sample
  means is lower ?x ? / ?n.

42
Example

• Given the population of men has normally
  distributed weights with a mean of 172 lb and a
  standard deviation of 29 lb, a) if one man is
  randomly selected, find the probability that his
  weight is greater than 167 lb.b) if 12
  different men are randomly selected, find the
  probability that their mean weight is greater
  than 167 lb.

43
Calculating the z-values
X as one possible single outcome
X as one possible sample mean
x - ?
x - ? x
z
z
?
? x
but ?x ? and ?x ? / ?n
167 -172
167 -172
z
z
so
29
29 / ?12
z
-0.17
z
-0.60
44
Using z to lookup p, we find that if one man is
randomly selected, the probability that his
weight is greater than 167 lb. is 0.5675.
45
Using z to lookup p, we find that if 12 different
men are randomly selected, the probability that
their mean weight is greater than 167 lb is
0.7257.
The frequency distribution of the sample means is
narrower (less variation) and taller than the
frequency distribution of the population but the
mean is the same.

46
The sample mean distribution has less dispersion
than the distribution of individual values

• Since the individual weights are more spread out
  than the sample average weights, only 56 of the
  area is under the curve for individual weights
  while 72 of the area is under the curve for
  average weights.
• For example, while an outlier can have a big
  effect on the distributions variation for
  individual weights, when only plotting sample
  averages, an outlier will get averaged into other
  values and will not be so outlying.
• Sample means cluster together more than
  individual values so it is more unusual for group
  of 12s average value to deviate from the mean
  than it is for an individual value to deviate
  from the mean.

47
Practical Interpretation

• There is a .5675 probability that an individual
  man will weight more than 167 lbs and there is a
  .7257 probability that 12 men will have a mean
  weight of more than 167 lbs.
• Given that the gondola maximum capacity is 2004
  lbs, it is likely (.72) to be overloaded if it is
  filled with 12 randomly-selected men.
• However, there is some hope that the gondola
  carriage wont come crashing down to earth
  because (1) skiers are generally leaner than the
  general public, (2) some of the 12 passengers are
  likely to be women, and (3) 2004 lbs is a very
  conservative limit in reality it can hold a lot
  more weight than that.

48
Applying the Central Limit Theorem when
hypothesis testing.

• So far, the only question we have learned how to
  answer is this What is the probability that x is
  x when ? is ? and ? is ??
• How useful is that? In each case, we already knew
  the population mean (and std. deviation) so why
  was it so helpful to know the probability of
  getting a certain x when ? was known already?
• In a more realistic situation, we hypothesize
  about what ? is, and we use the value of x (a
  sample mean) to determine whether to accept or
  reject that hypothesis.

49
How we use all this x, z, and p stuff to do
hypothesis testing

• A typical hypothesis I believe that the mean of
  this population is x.
• Written like this Ho (? x)
• And the way that hypothesis is tested is by
  taking a sample. Now if the mean of that sample
  is a lot different than the mean you are claiming
  for the population (x), you reject the
  hypothesis.

50
a lot different?

• The Rare Event Rule If, under a given assumption
  (? x), the probability of getting a particular
  x is really small (that is x and ? are really far
  apart the z-value is large), we conclude that
  the assumption (? x) is probably false (and
  reject the hypothesis).

51
Example

• Assume that the population of human body
  temperatures has a mean of 98.6oF as is commonly
  believed. Also assume that the population std.
  deviation is .62oF. If a sample of size n106 is
  randomly selected, and its mean turns out to be
  98.2oF, should we still hang on to the belief
  that the population mean is 98.6?

52
Assumptions we can make because of the Central
Limit Theorem

• We imagine that our sample of 106 is one of many
  possible samples of size 106 and if all those
  samples were taken and means obtained, we can
  assume that
• the distribution of those sample means would be
  normal (since n30),
• their mean of means ?x would be the same as the
  population mean (? 98.6)
• an their std deviation (?x) would be ? / ?n
  (or .62 / ?106 or .0602.

53
Hypothesis Ho (? 98.6)

• Based on these assumptions, we can calculate a z
  value using ?x 98.6, ?x .0602.
• From z we derive p, and if p is the hypothesis that the mean of the population is
  98.6

Since the p-value associated with a z of -6.64 is
off the charts, the probability is like
.00000000002 (clearly less than 5), so we reject
the hypothesis that ? 98.6
98.2 -98.6
z
.62 / ?106
z
-6.64
54
Practice exercise (p. 268- 9)

• The Rock-n-Roller Coaster at Disney-MGM Studio in
  Orlando has two seats in each row. When designing
  that roller coaster, the total width of the two
  seats in each row had to be determined. In the
  worst case scenario, both seats are occupied by
  men. Men have hip breadths that are normally
  distributed with a mean of 14.4in. And a standard
  deviation of 1.0 in. Assume that two male riders
  are randomly selected.
• A. Find the probability that their mean hip width
  is greater than 16.0 in.
• B. If each row of two seats is designed to fit
  two men only if they have a breadth of 16.0
  inches or less, would too many riders be unable
  to fit? Does this design appear to be acceptable?

55
This question is whats the probability of
getting a sample with a mean hip width of x?
NOT whats the probability of getting an
individual with hip width x?
56
Can we apply the central limit theorem
assumptions?

• The sample size is 2. Since the sample size (n) 30, we cannot conclude that the distribution of
  sample means will be a normal distribution.
• If we cant conclude that, then the central limit
  theorem assumptions (?x ? and ?x ? / ?n)
  cant be made.
• But since they told us that the underlying
  population has a normal distribution, we know
  that the sample means distribution is also
  normal, regardless of what n is.

57
Draw pictureWhat is P(x 16) when ?x 14.4?
The red curve line is the distribution of the
sample means. The blue curve is the distribution
of the population.
What is p?
?x 1.0/?2
? 1.0
X16
? ?x 14.4
58
2. Calculate z
x - ? x
z
? x
but ?x ? and ?x ? / ?n
16 -14.4
1.6
2.26
z
so

1.0 / ?2
.707
59
Lookup p
60
P(x 16) 1 - .9881 .0119
P is .0119
?x 1.0/?2
? 1.0
X16
? ?x 14.4
61
Is this acceptable?

• If each row of two seats is designed to fit two
  men only if they have a breadth of 16.0 inches or
  less, would too many riders be unable to fit?
  Does this design appear to be acceptable?
• About 1 of each sample of 2 men wont fit into
  the seats. Yes, that appears to be acceptable
  since really fat people would probably pass out
  waiting in line to get on the ride anyway.

62
Practice exercise (p. 269 - 16)

• Scores of men on the verbal portion of the SAT-I
  test are normally distributed with a mean of 509
  and a standard deviation of 112.
  Randomly-selected men are given the Columbia
  Review Course before taking the SAT test. Assume
  that the course did not help improve their scores
  (the null hypothesis).
• A. If 1 of the men are randomly selected, find
  the probability that his score is at least 590.
• B. If 16 men are randomly selected, find the
  probability that their mean score is at least
  590.
• C. In finding the probability for part (b), why
  CAN the central limit theorem be used even though
  the sample size is below 30?
• D. If the random sample of 16 men does result in
  a mean score of 590, is there strong evidence to
  support the claim that the course is actually
  effective? Why or why not?

63
Find P(x) 590

• A. If 1 of the men is randomly selected, find the
  probability that his score is at least 590.

64
1. Draw pictureWhat is P(x 590)?
What is p?
? 112
? 509
X590
65
2. Calculate z
x - ?
z
?
590 - 509
z
112
z
.723
66
3. Use z to lookup p
67
4. Revisit picture and write in pP(x 590) 1 -
. 7642 .2358
P .2358
? 112
? 509
x590
68
Find P(x) 590

• B. If 16 men are randomly selected, find the
  probability that their mean score is at least
  590.
• This question is whats the probability of
  getting this particular sample average? NOT
  whats the probability of getting this
  particular individual value?

69
1. Draw pictureWhat is P(x 590)?
What is p?
?x 112/?12
? 112
X590
? ?x 509
70
2. Calculate z value
x - ? x
z
? x
590 -509
81
2.89
z

112 /?16
28
71
3. Use z to lookup p
2.8
.9981
72
4. Revisit picture and write in pP(x) 590 1
- .9981 .0019
P .0019
?x 112/?16
? 112
X590
? ?x 509
73
C. In finding the probability for part (b), why
CAN the central limit theorem be used even though
the sample size is below 30?

• Because the underlying population (all SAT
  scores) is normal.

74
D. If the random sample of 16 men does result in
a mean score of 590, is there strong evidence to
support the claim that the course is actually
effective? Why or why not?

• Yes, because it is very unusual (.0019 16 men to obtain an average score that high.

75
Lets summarize

• The first part of the chapter was about Whats
  the probability of getting an outcome (x) that is
  less than or greater than some value? and Is it
  normal to get an outcome like this?
• The second part of this chapter applied these
  question to situations where the outcome variable
  was not a single individual outcome but an
  average of a sample of outcomes. In other words,
  Whats the probability of getting this
  particular sample average? and Is it normal to
  get such as sample average if the population mean
  is really what we think it is?

76
Homework 12

• 4 on page 263
• Random sampling from finite population
• 10 on page 264
• Finite or infinite populations?
• 16 on page 267
• Develop a point estimate
• 26 on page 278
• Whats the probability that sample statistic is
  close to population parameter?
• 38 on page 284
• Whats the probability that sample statistic is
  close to population parameter?