Statistics 221

1
Statistics 221

• Chapter 8
• Interval Estimation

2
What this chapter is about

• This chapter is about using sample statistics to
  infer population parameters.
• We learned in chapter 7 that we can use sample
  statistics (such as sample means and proportions)
  as point estimates (best guesses) of population
  parameters.
• But point estimates should not be treated as
  precise values of population parameters.
• Instead, it should be acknowledged that there is
  a some possibility for inaccuracy when using a
  sample statistic to drawn inferences about a
  population. To compensate for that
  possibility, statisticians create interval
  estimates.
• In other words, instead just claiming that ? x,
  we state that ? probably falls within an
  interval of xs.
• This chapter is about developing such interval
  estimates.

3
Interval estimates

• We provide interval estimates by taking a point
  estimate and adding and subtracting a value
  called the margin of error.
• For example, if we take a survey and find that
  our sample mean x is 25, we dont claim that
• ? 25.
• Instead we compute a margin of error value
  (lets say its 3) and we can claim that
• The true ? is probably somewhere between 25-3
  (22) and 253 (28).

4
Developing an interval estimate for a population
mean when ? is known

• When the population standard deviation ? is
  known, the formula for developing an interval
  estimate is different than if ? were not known.
• Realistically, if ? were known, then so would ?
  (since we use ? in the formula to calculate ?)
  and it wouldnt be necessary to use a sample
  statistic x to estimate ?.
• But the ? known situation is presented first so
  that it can be compared to the ? not known
  situation which we will examine next.

5
Example 1 Develop Confidence Interval for ? (?
known)

• A company named CJW, Inc. conducts a customer
  survey each month to monitor customer
  satisfaction with its products and services.
• They send out surveys on a monthly basis and ask
  customers about their satisfaction with such
  things as ease of placing orders, timely
  delivery, price, etc. They use the questionnaire
  results to develop an overall satisfaction
  score for each survey participant.
• A mean satisfaction score is derived from the
  sample and it provides a point estimate of the
  populations satisfaction score. But we use the
  point estimate to develop an interval estimate of
  what we think is the populations satisfaction
  score.

6
1. Identify/Obtain inputs

• Lets say that based on previous surveys, CJW,
  Inc knows that ?20.
• Lets say that a survey is taken of 100 customers
  and their overall satisfaction score is 82 (out
  of 100).
• Now we know that
• n100
• x 82
• ?20.

7
2. Verify assumptions regarding the distribution
of means

• We visualize this one sample of 100 as one of
  all the potential samples of size 100 that could
  be taken out of this population. From this one
  sample we calculate a mean which is one mean out
  of all potential sample means (where n100).
• And since n gt 30, we know that if we make a
  frequency distribution of all potential sample
  means, it will be normally distributed.
• Further, the mean of this distribution of sample
  means is equal to the population mean.
• Further, since we know that ? 20, the standard
  deviation of this distribution of sample means
  is

?
20

2

?x
v n
v100
8
3. Specify a Confidence Coefficient

• We dont actually say
• The true ? is probably somewhere between 25-3
  (22) and 253 (28).
• What we do say
• We are x confident that the true ? is somewhere
  between 25-3 (22) and 253 (28).
• That x is called the confidence coefficient and
  it is usually set at 99, 95 or 90 (you
  decide).
• Lets say we use 95 for this example.

9
The confidence interval

• The higher the confidence coefficient, the wider
  would be our confidence interval.
• For example
• If the confidence coefficient was 90, the
  confidence interval might be 24-26.
• If the confidence coefficient was 95, the
  confidence interval might be 23-27.
• If the confidence coefficient was 99, the
  confidence interval might be 22-28.

10
What the confidence coefficient is

• It turns out that this sample had a mean of 82.
  But we acknowledge that if we took another sample
  of 100, we would probably get a mean of something
  different (e. g., 81, 83, 80, etc).
• An interval estimate can be calculated from each
  of these point estimates and each of these
  intervals would be slightly different.
• In fact, if there were (N)! potential samples of
  100, then there are (N)! possible interval
  estimates.
• The confidence coefficient is the percentage of
  those interval estimates that contain the true
  population parameter ?.

11
Confidence Intervals from numerous samples.
We want to specify a width for our confidence
interval so that 95 of all possible confidence
intervals will contain the populations true
value.
x86
x85
x84
x83
TRUE POPULATION VALUE
x.82
x81
x80
x79
x78
12
Calculate a width of the confidence interval

• The next thing we must do is to calculate a
  width for our confidence interval so that 95 of
  all possible confidence intervals will contain
  the populations true value.
• How wide must that be?
• The answer will (at first) be expressed as a
  z-value.

Confidence Coefficent ? 95 ?
82
13
4. Calculate z?/2

• If the confidence coefficient is 95, we are
  leaving out 5 of the samples
• 2.5 in each of the two tails.
• To express that percentage as distance in zs
• We define alpha (?) as 1 - .95
• ? .05
• ? / 2 .025
• Using Excels normsinv(.025 ) z -1.96
• The width of the confidence interval should
  extend 1.96 standard deviations in each direction
  from the mean to encompass 95 of the area of the
  distribution.

14
If Confidence Coefficient .95, Then ? .05,
z 1.96
? This width encompasses 95 of all sample means
?
?/2 .025
?/2 .025
Z - 1.96
Z 1.96
This is the distribution of sample means
15
5. Calculate the Margin of Error (E)

• z is then multiplied by the standard deviation
  of this distribution of sample means to get an
  exact value for the margin of error.

(20)
?
E z ?/2
1.96
?100
?n
3.92
16
6. Calculate the confidence interval

• The margin of error (E) is both added and
  subtracted from the mean to calculate the
  confidence interval.

(82 3.92) lt ? lt (82 3.92) 78.08 lt ? lt
85.92

• Conclusion we are 95 confident that the true
  population mean ? falls between 78.08 and 85.92.

17
Example 2Develop Confidence Interval for ? (?
known)

• In order to help identify baby growth patterns
  that are unusual, we need to construct a
  confidence interval estimate of the mean head
  circumference of all babies that are two months
  old. A random sample of 100 babies is obtained,
  and the mean head circumference is found to be
  40.6 cm. Assuming that the population ? is known
  to be 1.6 cm, find a 99 confidence interval
  estimate of the mean head circumference of all
  two-month-old babies.

18
1. Identify/Obtain inputs

• n 100
• x 40.6 cm
• ? 1.6 cm.

19
2. Verify assumptions regarding the distribution
of means

• Since n gt 30, we know that if we make a frequency
  distribution of all potential sample means, it
  will be normally distributed.
• Further, the mean of this distribution of sample
  means is equal to the population mean.
• Further, since we know that ? 1.6, the standard
  deviation of this distribution of sample means
  is

?
1.6

.16

?x
v n
v100
20
3. Specify a Confidence Coefficient

• A 99 confidence coefficient has been set.

21
4. Calculate z?/2

• ? 1 - .99
• ? .01
• ?/2 .005
• z?/2 normsinv(.005) -2.576

22
5. Calculate the Margin of Error (E)

• z is then multiplied by the standard deviation
  of this distribution of sample means to get an
  exact value for the margin of error.

(1.6)
?
E z ?/2
2.576
?100
?n
.4
23
6. Calculate the confidence interval

• The margin of error (E) is both added and
  subtracted from the mean to calculate the
  confidence interval.

(40.6 .4) lt ? lt (40.6 .4) 40.2 lt ?
lt 41.0

• Conclusion we are 99 confident that the true
  population mean ? falls between 40.2 and 41.0.

24
Example 3Develop a Confidence Interval for ?
(? known)

• The health of the bear population in Yellowstone
  National Park is monitored by periodic
  measurements taken from anesthetized bears. A
  sample of 54 bears has a mean weight of 182.9.
  Assume that ? is known to be 121.8 lbs, find a
  99 confidence interval estimate of the mean of
  the population of all such bear weights.

25
1. Identify/Obtain inputs

• n 54
• x 182.9 lbs
• ? 121.8 lbs.

26
2. Verify assumptions regarding the distribution
of means

• Since n of 54 gt 30, we know that if we make a
  frequency distribution of all potential sample
  means, it will be normally distributed.
• Further, the mean of this distribution of sample
  means is equal to the population mean.
• Further, since we know that ? 121.8, the
  standard deviation of this distribution of sample
  means is

?
121.8


16.57
?x
v n
v54
27
3. Specify a Confidence Coefficient

• A 99 confidence coefficient has been set.

28
4. Calculate z?/2

• ? 1 - .99
• ? .01
• ?/2 .005
• z?/2 normsinv(.005) -2.576

29
5. Calculate the Margin of Error (E)

• z is then multiplied by the standard deviation
  of this distribution of sample means to get an
  exact value for the margin of error.

(121.8)
?
E z ?/2
2.576
?54
?n
42.68 lbs.
30
6. Calculate the confidence interval

• The margin of error (E) is both added and
  subtracted from the mean to calculate the
  confidence interval.

182.9 42.68) lt ? lt (182.9 42.68) 140.22
lt ? lt 225.58

• Conclusion we are 99 confident that the true
  population mean ? (weight of bears) falls between
  140.22 lbs. and 225.58 lbs.

31
Estimating a Population Mean ? when ? is unknown

• In the previous section, we estimated the
  population mean when ? was known. But that
  scenario would be unrealistic because we need the
  ? to calculate ?.
• Again, the scenario involves obtaining one sample
  from a population of samples and the
  distribution of those sample means is normal
  because either the underlying population
  distribution is normal or the sample size (n) is
  gt30.

32
when ? is unknown

• When ? is not known, we must estimate ? from s
  (the sample std deviation) and this introduces an
  additional source of error into the calculation
  of an interval estimate.
• To compensate for that source of error, we have
  to increase our margin of error which is going to
  widen our interval estimate.
• For example, if our margin of error was /- 3 if
  ? were known, it might be /- 4 when ? is
  unknown.
• So if the point estimate of a mean is 25, the
  interval estimate (at 95) might be
• (25-4) 21 lt ? lt 29 (254)

33
The distribution of sample means when ? is
unknown

• When ? is unknown, we cannot assume our
  distribution of sample means resembles the
  standardized (z) distribution.
• Instead we compensate for the additional source
  of uncertainty by assuming that our distribution
  of sample means resembles the (wider)
  t-distribution.

34
The t distribution wider than z
z
The way that the t-distribution works is the
smaller the sample size (n), the wider the curve
line.
35
The margin of error formula (when using the t
distribution)

• When ? is unknown, the t-value is used instead of
  the z-value.

?
s
E z ?/2
E t ?/2
?n
?n
36
Calculating a t-value

• The way that the t-distribution works is the
  smaller the sample size (n), the wider the curve
  line.
• Excels TINV( ) formula can be used but it
  requires two arguments
• tinv(?, n-1)
• The first argument is ? which is calculated as
  1-confident coefficient.
• The second argument is the degrees of freedom
  which is calculated as n-1.

37
The bears weight example

• In that example, ? was known to be 121.8. With a
  99 confidence coefficient, z was 2.576
• z normsinv(?/2)
• z normsinv(.005) 2.576
• If ? had been unknown but it was estimated with s
  121.8, then t would have been
• t tinv(?, n-1)
• t tinv(.01, 54-1) 2.672
• Now if n had been lower (say 30), t would have
  been
• t tinv(.01, 30-1) 2.756

Notice that ? is not divided by 2
38
Which distribution to use?
39
Which distribution should be used?

• 2. If n 10, ? is unknown, population appears to
  be normally distributed.
• 4. If n 45, ? is known, population appears to
  be very skewed.
• 6. If n 9, ? is known, population appears to be
  very skewed.
• 8. If n 37, ? is unknown, population appears to
  be normally distributed.

40
Example 1Developing a confidence interval for ?
(? unknown)

• A study was conducted to estimate hospital costs
  for accident victims who were wearing seat belts.
  Twenty randomly-selected cases have a
  distribution that appears to be bell-shaped with
  a mean of 9004 and a standard deviation of
  5629.
• A. Construct the 99 confidence interval for the
  mean of all such costs.
• B. If you are a manager for an insurance company
  that provides lower rates for drivers who wear
  seat belts and you want a conservative estimate
  for a worse case scenario, what amount should you
  use as the possible hospital cost for an accident
  victim who wears seat belts?

41
1. Identify/Obtain inputs

• n 20
• x 9004
• ? is estimated by s 5629

42
2a. Which distribution should you use?

• ? is unknown
• n 20
• But we are told that the population has a
  distribution which appears to be bell-shaped so
  well assume that the underlying population is
  normal, so the distribution of means is normal
  even though n lt 30.
• So use the t distribution

43
2b. Verify assumptions regarding the distribution
of means

• We have already concluded that since the
  underlying population is normal, the distribution
  of means is normally distributed.
• Further, the mean of this distribution of sample
  means is equal to the population mean.
• Further, since we know that ? is estimated by s
  5629, the standard deviation of this
  distribution of sample means is

?
5629


1258.68
?x
v n
v20
44
3. Specify a Confidence Coefficient

• A 99 confidence coefficient has been set.

45
4. Calculate t?/2

• ? 1 - .99
• ? .01
• t?/2 tinv(.01, 20-1) 2.861
• (Notice that the first argument, ? is not divided
  by 2 as it is with the normsinv( ) formula. The
  tinv( ) formula will automatically divide ? by 2.)

46
5. Calculate the Margin of Error (E)

• t is then multiplied by the standard deviation
  of this distribution of sample means to get an
  exact value for the margin of error.

(5629)
?
E t?/2
2.861
?20
?n
3601.01 lbs.
47
6. Calculate the confidence interval

• The margin of error (E) is both added and
  subtracted from the mean to calculate the
  confidence interval.

(9004 3601) lt ? lt (9004 3601) 5,403 lt
? lt 12,605

• Conclusion we are 99 confident that the true
  population mean ? (average medical cost of a car
  crash victim who was not wearing a seat belt)
  falls between 5,403 and 23,605.

48
Question B

• B. If you are a manager for an insurance company
  that provides lower rates for drivers who wear
  seat belts and you want a conservative estimate
  for a worse case scenario, what amount should you
  use as the possible hospital cost for an accident
  victim who wears seat belts?
• 12,605

49
Determining Sample Size

• In the previous examples, we had a pre-determined
  sample size and we used the sample size along
  with the mean, standard deviation, and confidence
  coefficient to calculate a margin of error and
  then a confidence interval estimate of a
  population parameter (?) based on a sample
  statistic (x).
• What if we want to fix the margin of error to be
  no more than a certain amount? We can do that by
  making the sample size sufficiently large.
• In this section, we calculate how large the
  sample size has to be in order to achieve a
  certain (minimal) margin of error that we will
  use in our interval estimates.

50
The sample size formula

• Using this formula, we can achieve a desired
  margin of error at a chosen confidence level.

2
z?/2 ?
n
E

• E is the margin of error that we are willing to
  accept.
• z?/2 will reflect the chosen confidence level.
• ? may or may not be known. If not, estimate it
  from s or do a separate pilot study to get an
  estimate.

51
Example 1Determine sample size to estimate a
population mean

• Assume that we want to estimate the mean IQ score
  for the population of statistics professors. How
  many statistics professors must be randomly
  selected for IQ tests if we want a 95 confidence
  that the sample mean is within 2 IQ points of the
  population mean?
• Recall that the formula requires the population
  ?. We can either use an overly-conservative
  guestimate or use the samples std deviation (s).

52
Determine Sample Size

• The confidence coefficient is set at 95, so z?/2
  is 1.96.
• The margin of error we are willing to accept (E)
  2 IQ points.
• ? 15. (We know that 15 is the ? for the
  general population and so we use this as our
  overly-conservative estimate.)

2
2
z?/2 ?
1.96 (15)
n

? 217
E
2
53
Example 2Determine sample size to estimate a
population mean

• The Tyco Video Game Corporation finds that it is
  losing income because of slugs used in video
  games. The machines must be adjusted to accept
  coins only if they fall within set limits. In
  order to set those limits, the mean weight of
  quarters in circulation must be estimated. A
  sample of quarters will be weighed in order to
  determine the mean. How many quarters must we
  randomly select and weigh if we want to be 99
  confident that the sample mean is within .025
  grams of the true population mean for all
  quarters? Based on results from the sample of
  quarters, we estimated the population standard
  deviation to be .068 g.

54
Determine Sample Size

• Desired Confidence Coefficient 99
• Desired Margin of Error (E) .025 grams
• Population ? .068 grams

2
2
z?/2 ?
2.575 (.068)
n

? 50
E
.025
55
Estimating a population proportion (P)

• A proportion is a percentage expressed as a
  decimal value.
• A proportion (P) is a way of referring to to the
  percentage of cases (e. g. people or subjects)
  that have some characteristic or opinion (the
  successes).
• For example, we may want to make inferences about
  the proportion of people who favor gun control,
  support the death penalty, use email, or that
  watched a particular TV show.
• In these situations, we are trying to discover
  what P is, where P is the percentage/proportion
  of people who posses the characteristic or
  opinion of interest.

56
P is the random variable

• In this scenario, P the proportion of the
  population (of size N) that have this
  characteristic or opinion (the successes) is
  the unknown.
• In the last chapter, P, the probability of having
  a girl was known (50) and the P(having x girls)
  was the unknown. In this scenario, P, the
  probability / proportion is the unknown.
• Theoretically speaking, if we took all possible
  samples of size n, and got a p for each sample,
  and created a probability distribution of ps,
  that distribution would have an average value of
  P.
• In reality, we take one sample, and get one p and
  then draw inferences about what P is from that
  one p.

57
Estimating a population proportion

• Here is the scenario 829 people who live in
  Minnesota were asked if they were in favor of
  using a photo-cop system that uses cameras to
  ticket drivers.
• 51 of the 829 said they were opposed to it. So
  51 what we call the sample proportion (denoted
  by p).
• Can we say that the proportion of all Minnesotans
  (P) that oppose photo-cop is 51?

58
Formula for the Margin of Error (E)(when
estimating a proportion)

(.51) (.49)
(p)(q)
E z?/2
1.96
n
829
.0345
59
We can now calculate the Confidence Interval

• The confidence interval .51 /- .034
• OR
• (p E) lt P lt (p E)
• .476 lt P lt .544
• We are 95 sure that the true population
  parameter (P) is between 47.6 and 54.4

60
Determining Sample Size when estimating a
population proportion

• How do you know how large of a sample you need
  when gathering your data?
• Its not just a matter of making it large enough
  so that the sampling distribution is normal (the
  n gt 30 rule).
• Its also a function of what margin of error you
  are willing to accept (e.g. .03, .04, etc.)
• And what confidence level you chose (e. g., 90,
  95, 99).

61
The sample size formula for estimating population
proportions


z ?/2 2 (p)(q)
n
E2

• As you can see, it incorporates the margin of
  error (E), the confidence level (1 - ?), and the
  product of (p) (q).

62
Calculating Sample Size when we have an estimated
p

• If we have some prior information suggesting what
  p is, we use this formula

z ?/2 2 (p)(q)
n
E2
63
Calculating Sample Size when we do NOT have an
estimated p

• If we have NO prior information suggesting what p
  is, we use the value that maximizes p q (.5
  .5 .25)

z ?/2 2 (.25)
n
E2
64
Sample size for an Email Survey

• Suppose you want to do a survey to determine the
  current percentage of US households that use
  email.
• How many households must be surveyed in order to
  be 95 confident that the sample percentage is in
  error by no more than 4.
• A. In an earlier study (1997), the percentage was
  16.9. Use this for your p.
• B. Assume that we have no prior information
  suggesting a possible value for p.

65
A. Assume p .169


z ?/2 2 (p)(q)
1.96 2 (.169)(.831)
n

E2
.042
337.194 (round to 338)
66
b. Assume p is unknown
z ?/2 2 (.25)
1.96 2 (.25)
n

E2
.042
600.25 (round to 601)
When p of .5 is used, the required sample size
almost doubles.
67
Sample size ? of population size

• The population size is not used in the formula to
  determine sample size.
• Generally, this is the case unless we are
  sampling a small population without replacement.
• Neilson, for example surveys 4,000 TV households
  from a population of 104 million households
  (.004) and can still be 95 confident that the
  sample p will be within 1 of the population P.

68
Use the Confidence Intervals to find the point
estimate p and the margin of error E. Practice
exercise (p. 312)

• 10. (.278 lt p lt .338)
• 12. (.887 lt p lt .927)

69
Find the point estimate p and the margin of error
E.

• 10. (.278 lt p lt .338)

.278 .338
p
.308
2
.338 - .278
E
.03
2
70
Find the point estimate p and the margin of error
E.

• 12. (.887 lt p lt .927)

.887 .927
p
.907
2
.927 - .887
E
.02
2
71
Find the margin of error E that corresponds to
the given statistics and confidence
levelPractice exercise p. 312

• 14. n 1200, x 400, CL 99
• 16. CL 95, sample size 500, 80 are successes

72
Find the margin of error E that corresponds to
the given statistics and confidence level

• 14. n 1200, x 400, CL 99
• If CL 99, then z 2.575
• If x 400 and n 1200, then p .33, q .67

(.33) (.67)
(p)(q)
E z ?/2
2.575
n
1200
.0350
73
Find the margin of error E that corresponds to
the given statistics and confidence level

• 16. 95 CL, sample size is 500, 80 are
  successes.
• If CL 95, then z 1.96
• p .80, q .20, n 500

(.80) (.20)
(p)(q)
E z ?/2
1.96
n
500
.0351
74
Construct the confidence interval estimate of the
population proportion PPractice Exercise (P. 312)

• 18. n 1200, x 200, CL 99
• 20. n 2001, x 1776, CL 90

75
Construct the confidence interval estimate of the
population proportion P

• 18. n 1200, x 200, CL 99
• p 200/1200 or .167, q .833, z 2.575

(.167) (.833)
(p)(q)
E z ?/2
2.575
n
1200
.028
(.167 .028) lt P lt (.167 .028) .139 lt
P lt .194
76
Construct the confidence interval estimate of the
population proportion P

• 20. n 2001, x 1776, CL 90
• p 1776/2001 .887, q .113, z 1.645

(.887) (.113)
(p)(q)
E z ?/2
1.645
n
2001
.0116
(.887 .0116) lt P lt (.887 .0116) .876 lt
P lt .899
77
Use the given data to find the minimum sample
size to estimate a population proportion or
percentagePractice Exercise (p. 312)

• 22. E .038, CL 95, p and q unknown
• 24. E .03, CL 90, estimated p .08

78
Use the given data to find the minimum sample
size to estimate a population proportion or
percentage

• 22. E .038, CL 95, p and q unknown

z ?/2 2 (p)(q)
1.96 2 (.25)
n

E2
.0382
665.10 (round to 666)
79
Use the given data to find the minimum sample
size to estimate a population proportion or
percentage

• 24. E .03, CL 90, estimated p .08

z ?/2 2 (p)(q)
1.645 2 (.08)(.92)
n

E2
.032
221.29 (round to 222)
80
Practice Exercise (p. 313 - 28)

• In a Gallup poll, 491 randomly-selected adults
  were asked whether they are in favor of the death
  penalty for a person convicted of murder, and 65
  said they were in favor of the death penalty.
• A. Find the point estimate of the percentage of
  adults who are in favor of the death penalty.
• B. Find a 95 confidence interval of the
  percentage of adults who are in favor of the
  death penalty.
• C. Can we safely conclude that the majority of
  adults are in favor of this death penalty?
  Explain.

81
A B Find the point estimate and confidence
Interval of the percentage of adults who are in
favor of the death penalty.

• A. The point estimate is 65
• B. The confidence interval is

(.65) (.35)
(p)(q)
E z ?/2
1.96
n
491
.04
(.65 .04) lt P lt (.65 .04) .61 lt P lt
.69
82
C. Can we safely conclude that the majority of
adults are in favor of this death penalty?
Explain.

• Yes, since the entire interval is above 50.

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Homework 13

• 6 on page 304
• develop confidence interval
• 18 on page 313
• develop confidence interval
• 28 on page 316
• sample size
• 38 on page 321
• confidence interval and sample size