Outline

1
Outline

• Review of last week
• Sampling distributions
• The sampling distribution of the mean
• The Central Limit Theorem
• Confidence intervals
• Normal distribution example
• Sampling distribution example
• Confidence interval example

2
Review of last week

• Last week, we learned how to use the Standard
  Normal Distribution to work out the probability
  of finding individual scores in some interval
  e.g., what is the probability that the next
  Canadian woman we meet is taller than 175 cm?
• Today, were going to do the same sort of thing
  with sample means rather than individual scores.

3

• The sampling distribution of a sample statistic
  (such as X) is the probability distribution of
  that statistic.

4

• The sampling distribution of the mean consists of
  all possible sample means for all possible
  samples of size n that you could take from the
  population

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Distribution of sample means for samples of size n
µX

• When we draw a sample from a population, we are
  at the same time drawing a sample mean from the
  distribution of sample means for samples of size n

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Sampling distributions

• The sampling distribution of a sample statistic
  is the probability distribution of that
  statistic.
• We can have sampling distributions of any sample
  statistic

• Mean
• Median M
• Variance s2
• Std devn s

7
The sampling distribution of the mean

• The sampling distribution of the sample mean X.
• E(X) µ µ
• Variability of this distribution is given by the
  standard error of the mean
• s s ? s

8
The Central Limit Theorem

• Consider a random sample of n observations from a
  population with mean µ and standard deviation ?.

• When n is sufficiently large, the sampling
  distribution of X will be approximately normal
  with mean µ µ and ? ?/ .
• Note this is true regardless of the shape of the
  underlying distribution of raw scores

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The Central Limit Theorem

• The larger the sample size, the better the
  approximation to the normal distribution.

• For most populations, n 30 will be
  sufficiently large.

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The Central Limit Theorem

• When we draw a sample and measure its mean, by
  the CLT, we may assume the sampling distribution
  of the sample mean is normal.
• That means we can use the standard normal
  distribution (SND) to work out the probability of
  finding a sample mean in a given range relative
  to the population mean.

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?
µ
The sampling distribution of the sample mean
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The sampling distribution of the mean

• We use the sampling distribution of the mean the
  way we used the SND last week. We obtain
  probabilities of finding sample means in a given
  range relative to the population mean, for
  samples of size n.
• Dont forget to use the standard error, sX,
  rather than the standard deviation, s!

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Confidence Intervals

• There are two ways to estimate population
  parameters such as the mean
• Point estimates, such as X
• Interval estimates, which tell us a range of
  values that will contain the parameter with known
  probability.

14
.45
.45
Z -1.645 µX Z 1.645
90 of the time, X will fall within the range
Z -1.645 to Z 1.645
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Confidence Intervals

• If 90 of the time X falls in the range Z
  -1.645 to Z 1.645 around the mean µ, then
• 90 of the time, µ must fall within a range of
  the same width centered on X.

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Confidence Intervals

• For given ?, the 100 (1-?) Confidence Interval
  for µX is
• C.I. X Z?/2 ?X
• C.I. X Z?/2 ?/vn

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Confidence Intervals

• When ? is not known and n is large ( 30), use s
• C.I. X Z?/2 sX
• C.I. X Z?/2 s/vn

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Normal Distribution Example

• The amount of time that students wait to be
  served when buying coffee from the Campus Perks
  coffee outlet is normally distributed with a mean
  of 62.0 seconds and a 98.5 percentile of 79.36
  seconds. In a random sample of 30 students buying
  coffee at Campus Perks, approximately how many
  will wait between 40 and 58 seconds to be served?

NOTE This is not a question about a sample mean!
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Normal Distribution Example
.50
.4850
40 58 62 P98.5 Z for .4850 2.17
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Normal Distribution Example

• ? 79.36 62 8
• 2.17
• Z1 40 62 -2.75 (p .4970 from table)
• 8
• Z2 58 62 -0.50 (p .1915 from table)
• 8

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Normal Distribution Example

• P(40 X 58) .4970 - .1915 .3055
• The probability of any one student waiting
  between 40 and 58 seconds is .3055.
• Therefore, in a random sample of 30, we expect
  approximately .3055 (30) 9.165 9 students to
  wait between 40 and 58 seconds.

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Sampling Distribution Example

• Peoples reaction times (RTs) to a simple visual
  stimulus are normally distributed with a mean of
  500 milliseconds and a standard deviation of 150
  milliseconds. You believe that people who go on a
  low-carb diet, however, will have slower (longer)
  RTs than this, on average, though their standard
  deviation will remain at 150. To test your
  belief, you take a random sample of 40 people who
  self-report having being on a low-carb diet for
  at least 6 months and measure their RTs. You
  decide that your belief will be supported if the
  mean RT of the low-carb group is 565 milliseconds
  or slower. What is the probability that you will
  conclude that your belief has been supported even
  if a low-carb diet actually has no effect on RTs
  whatsoever?

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We want this probability
500 565
You decide that your belief will be supported if
the mean RT of the low-carb group is 565
milliseconds or slower. What is the probability
that you will conclude that your belief has been
supported even if a low-carb diet actually has no
effect on RTs whatsoever?
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Example 2

• What is P(X 565 µ 500)?
• Z 565 500
• 150/v40

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Example 2

• What is P(X 565 µ 500)?
• Z 565 500 65 2.74
• 150/v40 23.72
• P for Z 2.74 (from table) is .4969.
• Therefore, desired probability is .5 - .4969
  .0031.

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Example 3

• Two variables important to a professional
  football player are speed and strength. Each
  year, camps are held to determine potential
  players speed and strength, both of which are
  continuous, normally-distributed, and independent
  of each other. The middle 95 of strength scores
  is bounded by 600 and 900 (on a composite
  strength index). The average time to run 40 yards
  is 4.6 seconds, and 40 yard time exceeds 6
  seconds only 5 of the time.
• a. In order to be considered by a team, a
  potential player must not exceed the 75th
  percentile for time to run 40 yards. What is the
  slowest a player can run 40 yards and still be
  considered?

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.45
seconds
X
Probability distribution for time to run 40 yards
(seconds)
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Example 3

• Z(.45) 1.645 6 4.6
• s
• s 6 4.6 .851
• 1.645

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Example 3

• Now we can find X (the 75th percentile)
• Z(.25) 0.675 X 4.6
• .851
• X 0.675 (.851) 4.6 5.15 (seconds)

30
seconds
5.15
The 75th percentile for 40 yard times is 5.15
seconds.
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Example 3

• Two variables important to a professional
  football player are speed and strength. Each
  year, camps are held to determine potential
  players speed and strength, both of which are
  continuous, normally-distributed, and independent
  of each other. The middle 95 of strength scores
  is bounded by 600 and 900 (on a composite
  strength index). The average time to run 40 yards
  is 4.6 seconds, and 40 yard time exceeds 6
  seconds only 5 of the time.
• b. You take a random sample of 200 potential
  players. What is the probability that the average
  strength score of the sample is less than or
  equal to 740?

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µ
750
Probability distribution for strength scores
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Example 3

• Z 1.645 900 750
• s
• s 900 750 91.19
• 1.645
• Z 740 750 -1.55
• 91.19/v200

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Example 3

• P (Z lt 1.55) .4394 (From table)
• Tail probability will be .5 .4394 .0606

35
750
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Confidence Interval Example

• A researcher samples 36 undergraduates from a
  local university and finds it took them 36.4
  days, on average, to find a job, with a standard
  deviation of 8 days. Use these data to form a 96
  confidence interval for the true mean time it
  takes for graduates to find a job.

NOTE We are not given the population standard
deviation
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Confidence Interval Example

• Recall
• C.I. X Z?/2 sX X Z?/2 s/vn
• X 36.4
• S 8
• n 36
• S/vn 8/6 1.33

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Confidence Interval Example

• (1-?) 96, so ?/2 .02 this is the tail
  probability.
• We get ?/2 .02 when we look up Z.48 2.05
• C.I. 36.4 2.05 (1.33)
• (33.67 µ 39.13)