Confidence Interval Estimation

1
Confidence Interval Estimation

• For statistical inference in
• decision making
• Chapter 6

2
Objectives

• Central Limit Theorem
• Confidence Interval Estimation of the Mean (s
  known)
• Interpretation of the Confidence Interval
• Confidence Interval Estimation of the Mean (s
  unknown)
• Confidence Interval Estimation for the Proportion
• Determining Sample Size

3
Central Limit Theorem

• Irrespective of the shape of the underlying
  distribution of the population, by increasing the
  sample size, sample means proportions will
  approximate normal distributions if the sample
  sizes are sufficiently large.

4
Central Limit Theorem in action
5
How large must a sample be for the Central Limit
theorem to apply?

• The sample size varies according to the shape of
  the population.
• However, for our use, a sample size of 30 or
  larger will suffice.

6
Must sample sizes be 30 or larger for populations
that are normally distributed?

• No. If the population is normally distributed,
  the sample means are normally distributed for
  sample sizes as small as n1.

7
Why not just always pick a sample size of 30?
8
How can I tell the shape of the underlying
population?

• CHECK FOR NORMALITY
• Use descriptive statistics. Construct
  stem-and-leaf plots for small or moderate-sized
  data sets and frequency distributions and
  histograms for large data sets.
• Compute measures of central tendency (mean and
  median) and compare with the theoretical and
  practical properties of the normal distribution.
  Compute the interquartile range. Does it
  approximate the 1.33 times the standard
  deviation?
• How are the observations in the data set
  distributed? Do approximately two thirds of the
  observations lie between the mean and plus or
  minus 1 standard deviation? Do approximately
  four-fifths of the observations lie between the
  mean and plus or minus 1.28 standard deviations?
  Do approximately 19 out of every 20 observations
  lie between the mean and plus or minus 2 standard
  deviations?

9
Why do I care if X-bar, the sample mean, is
normally distributed?
10
Because I want to use Z scores to analyze sample
means.

• But to use Z scores, the data must be normally
  distributed.
• Thats where the Central Limit Theorem steps in.
• Recall that the Central Limit Theorem states that
  sample means are normally distributed regardless
  of the shape of the underlying population if the
  sample size is sufficiently large.

11
Recall from Chapter 5

• Z (X - µ) s
• If sample means are normally distributed, the Z
  score formula applied to sample means would be
• Z X-bar - µX-bar s X-bar

12
Background

• To determine µX-bar, we would need to randomly
  draw out all possible samples of the given size
  from the population, compute the sample means,
  and average them. This task is unrealistic.
  Fortunately, µX-bar equals the population mean µ,
  which is easier to access.
• Likewise, computing the value of sX-bar, we would
  have to take all possible samples of a given size
  from a population, compute the sample means, and
  determine the standard deviation of sample means.
  This task is also unrealistic. Fortunately,
  sX-bar can be computed by using the population
  standard deviation divided by the square root of
  the sample size.

13
Note

• As the sample size increases,
• the standard deviation of the sample means
  becomes smaller and smaller
• because the population standard deviation is
  being divided by larger and larger values of the
  square root of n.

14
The ultimate benefit of the central limit theorem
is a useful version of the Z formula for sample
means.
15
Z Formula for Sample MeansZ X-bar - µ s
/ v n
16
Example

• The mean expenditure per customer at a tire store
  is 85.00, with a standard deviation of 9.00.
• If a random sample of 40 customers is taken,
  what is the probability that the sample average
  expenditure per customer for this sample will be
  87.00 or more?

17
Because the sample size is greater than 30, the
central limit theorem says the sample means are
normally distributed.

• Z X-bar - µ s / v n
• Z 87.00 - 85.00 9.00 / v 40
• Z 2.00 / 1.42 1.41

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• For Z 1.41 in the Z distribution table, the
  probability is .4207.
• This represents the probability of getting a
  mean between 87.00 and the population mean
  85.00.
• Solving for the tail of the distribution
  yields
• .5000 - .4207 .0793
• This is the probability of X-bar 87.00.

19
Interpretations

• Therefore, 7.93 of the time, a random sample of
  40 customers from this population will yield a
  mean expenditure of 87.00 or more.
• OR
• From any random sample of 40 customers, 7.93 of
  them will spend on average 87.00 or more.

20
Interpretations

• Therefore, 7.93 of the time, a random sample
  of 40 customers from this population will yield a
  mean expenditure of 87.00 or more.

• From any random sample of 40 customers, 7.93
  of them will spend on average 87.00 or more.

21
Solve

• Suppose that during any hour in a large
  department store, the average number of shoppers
  is 448, with a standard deviation of 21 shoppers.
• What is the probability that a random sample of
  49 different shopping hours will yield a sample
  mean between 441 and 446 shoppers?

22
Statistical Inference
23
Statistical Inference facilitates decision making.
24
Via sample data, we can estimate something
about our population, such as its average value
µ, by using the corresponding sample mean,
X-bar.
25
Recall that µ, the population mean to be
estimated, is a parameter, while X-bar, the
sample mean, is a statistic.
26
Point Estimate

• A point estimate is a statistic taken from a
  sample and is used to estimate a population
  parameter.
• However, a point estimate is only as good as the
  sample it represents. If other random samples
  are taken from the population, the point
  estimates derived from those samples are likely
  to vary.
• Because of variation in sample statistics,
  estimating a population parameter with a
  confidence interval is often preferable to using
  a point estimate.

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Confidence Interval

• A confidence interval is a range of values within
  which it is estimated with some confidence the
  population parameter lies.
• Confidence intervals can be one or two-tailed.

28
Confidence Interval to Estimate µ

• By rearranging the Z formula for sample means, a
  confidence interval formula is constructed
• X-bar /- Z a/2 s / v n
• Where
• a the area under the normal curve outside the
  confidence interval
• a/2 the area in one-tail of the distribution
  outside the confidence interval

29

• The confidence interval formula yields a range
  (interval) within which we feel with some
  confidence the population mean is located.
• It is not certain that the population mean is in
  the interval unless we have a 100 confidence
  interval that is infinitely wide, so wide that it
  is meaningless.

30
Confidence interval estimates for five different
samples of n25, taken from a population where
µ368 and s15
31
Common levels of confidence intervals used by
analysts are 90, 95, 98, and 99.
32
95 Confidence Interval

• For 95 confidence, a .05 and a / 2 .025. The
  value of Z.025 is found by looking in the
  standard normal table under .5000 - .025 .4750.
  This area in the table is associated with a Z
  value of 1.96.

• An alternate method multiply the confidence
  interval, 95 by ½ (since the distribution is
  symmetric and the intervals are equal on each
  side of the population mean.
• (½) (95) .4750 (the area on each side of the
  mean) has a corresponding Z value of 1.96.

33
In other words, of all the possible X-bar values
along the horizontal axis of the normal
distribution curve, 95 of them should be within
a Z score of 1.96 from the mean.
34
Margin of Error

• Z s / v n

35
Example

• A business analyst for cellular telephone company
  takes a random sample of 85 bills for a recent
  month and from these bills computes a sample mean
  of 153 minutes. If the company uses the sample
  mean of 153 minutes as an estimate for the
  population mean, then the sample mean is being
  used as a POINT ESTIMATE. Past history and
  similar studies indicate that the population
  standard deviation is 46 minutes.
• The value of Z is decided by the level of
  confidence desired. A confidence level of 95 has
  been selected.

36
153 /- 1.96( 46/ v 85) 143.22 µ 162.78

• The confidence interval is constructed from the
  point estimate, 153 minutes, and the margin of
  error of this estimate, / - 9.78 minutes.
• The resulting confidence interval is 143.22 µ
  162.78.
• The cellular telephone company business analyst
  is 95 confident that the average length of a
  call for the population is between 143.22 and
  162.78 minutes.

37
Interpreting a Confidence Interval

• For the previous 95 confidence interval, the
  following conclusions are valid
• I am 95 confident that the average length of a
  call for the population µ, lies between 143.22
  and 162.78 minutes.
• If I repeatedly obtained samples of size 85, then
  95 of the resulting confidence intervals would
  contain µ and 5 would not. QUESTION Does this
  confidence interval 143.22 to 162.78 contain µ?
  ANSWER I dont know. All I can say is that this
  procedure leads to an interval containing µ 95
  of the time.
• I am 95 confident that my estimate of µ namely
  153 minutes is within 9.78 minutes of the actual
  value of µ. RECALL 9.78 is the margin of error.

38
Be Careful! The following statement is NOT true

• The probability that µ lies between 143.22 and
  162.78 is .95.
• Once you have inserted your sample results into
  the confidence interval formula, the word
  PROBABILITY can no longer be used to describe the
  resulting confidence interval.

39
Confidence Interval Estimation of the Mean (s
Unknown)

• In reality, the actual standard deviation of the
  population, s, is usually unknown.
• Therefore, we use s (sample standard deviation)
  to compute the confidence interval for the
  population mean, µ.
• However, by using s in place of s, the
  standard normal Z distribution no longer applies.
• Fortunately, the t-distribution will work,
  provided the population we obtain the sample is
  normally distributed.

40
Assumptions necessary to use t-distribution

• Assumes random variable x is normally distributed
• However, if sample size is large enough ( gt 30),
  t-distribution can be used when s is unknown.
• But if sample size is small, evaluate the shape
  of the sample data using a histogram or
  stem-and-leaf.
• As the sample size increases, the t-distribution
  approaches the Z distribution.

41
Confidence Interval using a t-distribution

• X-bar /- t a,n-1 s / v n
• a confidence interval
• n-1 degrees of freedom

42
Example

• As a consultant I have been employed to estimate
  the average amount of comp time accumulated per
  week for managers in the aerospace industry.
• I randomly sample 18 managers and measure the
  amount of extra time they work during a specific
  week and obtain the following results (in hours).
  Assume a 90 confidence interval.
• AEROSPACE DATA
• 6 21 17 20 7 0 8 16 29
• 3 8 12 11 9 21 25 15 16

43
Solution

• To construct a 90 confidence interval to
  estimate the average amount of extra time per
  week worked by a manager in the aerospace
  industry, I assume that comp time is normally
  distributed in the population.
• The sample size is 18, so df 17.
• A 90 level of confidence results in an a / 2
  .05 area in each tail.
• The table t-value is t .05,17 1.740.

44

• With a sample mean of 13.56 hours, and a
  sample standard deviation of 7.8 hours, the
  confidence interval is computed
• X-bar /- t a/2, n-1 S / v n
• 13.56 /- 1.740 ( 7.8 / v 18) 13.56 /- 3.20
• 10.36 µ 16.76

45
Interpretation

• The point estimate for this problem is 13.56
  hours, with an error of /- 3.20 hours.
• I am 90 confident that the average amount of
  comp time accumulated by a manager per week in
  this industry is between 10.36 and 16.76 hours.

46
Recommendations

• From these figures, the aerospace industry could
  attempt to build a reward system for such extra
  work or evaluate the regular 40-hour week to
  determine how to use the normal work hours more
  effectively and thus reduce comp time.

47
Solve

• I own a large equipment rental company and I want
  to make a quick estimate of the average number of
  days a piece of ditch digging equipment is rented
  out per person per time. The company has records
  of all rentals, but the amount of time required
  to conduct an audit of all accounts would be
  prohibitive.
• I decide to take a random sample of rental
  invoices.
• Fourteen different rentals of ditch diggers are
  selected randomly from the files.
• Use the following data to construct a 99
  confidence interval to estimate the average
  number of days that a ditch digger is rented and
  assume that the number of days per rental is
  normally distributed in the population.

48
Ditch Digger Data

• 3 1 3 2 5 1 2 1 4 2 1 3 1 1

49
Stay-tuned
50
Estimating the Population Proportion

• For most businesses, estimating market share
  (their proportion of the market) is important b/c
  many company decisions evolve from market share
  information
• What proportion of my customers pay late?
• What proportion dont pay at all?
• What proportion of the produced goods are
  defective?
• What proportion of the population has cats/ dogs/
  horses/ kids/ exercises/ reads?

51
Confidence Interval Estimate for the Proportion

• ps /- Zv ps(1-ps) / n
• ps - Zvps(1-ps) /n p ps Zvps(1-ps) /n
• ps sample proportion X / n number of
  successes sample size. This is the POINT
  ESTIMATE.
• p population proportion
• Z critical value from the standardized normal
  distribution
• n sample size

52
ps /- Zv ps(1-ps) / n

• NOTE This formula can be applied only when np
  and n(1-p) are at least 5.

53
Example

• A study of 87 randomly selected companies with a
  telemarketing operation revealed that 39 of the
  sampled companies had used telemarketing to
  assist them in order processing.
• Using this information, how could a researcher
  estimate the population proportion of
  telemarketing companies that use their
  telemarketing operation to assist them in order
  processing?

54
Solution

• The sample proportion .39.
• This is the point estimate of the population
  proportion, p.
• The Z value for 95 confidence is 1.96.
• The value of (1-p) 1 - .39 .61.

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ps /- Zv ps(1-ps) / n

• ps - Zvps(1-ps) /n p ps Zvps(1-ps) /n
• The confidence interval estimate is
• .39 1.96v(.39) (.61) / 87 p .39
  1.96v(.39) (.61) / 87
• .39 - .10 p .39 .10
• .29 p .49

56
Interpretation

• We are 95 confident that the population
  proportion of telemarketing firms that use their
  operation to assist order processing is somewhere
  between .29 and .49.
• There is a point estimate of .39 with a margin of
  error of /- .10.

57
Solve

• A clothing company produces mens jeans. The
  jeans are made and sold with either a regular cut
  or a boot cut.
• In an effort to estimate the proportion of their
  mens jeans market in Oklahoma City that is for
  boot-cut jeans, the analyst takes a random sample
  of 212 jeans sales from the companys two
  Oklahoma City retail outlets.
• Only 34 of the sales were for boot-cut jeans.
• Construct a 90 confidence interval to estimate
  the proportion of the population in Oklahoma City
  who prefer boot-cut jeans.

58
Solution

• ps 34/212 .16
• A point estimate for boot-cut jeans is .16 or
  16.
• The Z value for 90 level of confidence is 1.645.
• The confidence interval estimate is
• ps - Zvps(1-ps) /n p ps Zvps(1-ps) /n
• .16 1.645v(.16) (.84) / 212 p .16
  1.645v(.16) (.84) / 212
• .16 - .04 P .16 .04
• .12 P .20
• We are 90 confident that the proportion of
  boot-cut jeans is between 12 and 20 .

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Estimating Sample Size

• The amount of sampling error you are willing to
  accept and the level of confidence desired,
  determines the size of your sample.

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Sample size when Estimating µ

• n Z2s2 / e2
• e Z (s / v n

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To determine sample size

• Know the desired confidence level, which
  determines the value of Z (the critical value
  from the standardized normal distribution.
  Determining the confidence level is subjective.
• Know the acceptable sampling error, e. The amount
  of error that can be tolerated.
• Know the standard deviation, s. If unknown,
  estimate by
• past data
• educated guess
• estimate s s range/4 This estimate is
  derived from the empirical rule stating that
  approximately 95 of the values in a normal
  distribution are within /- 2s of the mean,
  giving a range within which most of the values
  are located.

62
Example

• Suppose the marketing manager wishes to estimate
  the population mean annual usage of home heating
  oil to within /- 50 gallons of the true value,
  and he wants to be 95 confident of correctly
  estimating the true mean.
• On the basis of a study taken the previous year,
  he believes that the standard deviation can be
  estimated as 325 gallons.
• Find the sample size needed.

63
Solution

• With e 50, s 325, and 95 confidence (Z
  1.96)
• n Z2s2 /e2 (1.96)2 (325)2 / (50)2
• n 162.31
• Therefore, n 163. As a general rule for
  determining sample size, always round up to the
  next integer value in order to slightly over
  satisfy the criteria desired.

64
Solve

• Suppose you want to estimate the average age of
  all Boeing 727 airplanes now in active domestic
  U.S. service.
• You want to be 95 confident, and you want your
  estimate to be within 2 years of the actual
  figure.
• The 727 was first placed in service about 30
  years ago, but you believe that no active 727s in
  the U.S. domestic fleet are more than 25 years
  old.
• How large a sample should you take?

65
Solution

• With E 2 years,
• Z value for 95 1.96,
• and s unknown,
• it must be estimated by using s range 4. As
  the range of ages is 0 to 25 years, s 25 4
  6.25.

66
n Z2s2 /e2

• n Z2s2 /e2 (1.96)2 (6.25)2 / (2)2
• 37.52 airplanes.
• Because you cannot sample 37.52 units, the
  required sample size is 38.
• If you randomly sample 38 planes, you can
  estimate the average age of active 727s within 2
  years and be 95 confident of the results.

67
Solve

• Determine the sample size necessary to estimate µ
  when values range from 80 to 500, error is to be
  within 10, and the confidence level is 90 .
• n Z2s2 /e2
• Answer 200

68
Determining sample size for proportion

• n Z2p(1-p) /e2
• p population proportion (if unknown, analysts
  use .5 as an estimate of p in the formula)
• e error of estimation equal to (ps p) the
  difference between the sample proportion and the
  parameter to be estimated, p. Represents amount
  of error willing to tolerate.

69
Solve

• The Packer, a produce industry trade publication,
  wants to survey Americans and ask whether they
  are eating more fresh fruits and vegetables than
  they did 1 year ago.
• The organization wants to be 90 confident in
  its results and maintain an error within .05. How
  large a sample should it take?

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