Ch 8, Confidence

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Ch 8, Confidence

• In Ch7, we saw that the average of a sample has a
  normal distn
• The mean of the avg is the mean of the data
• We can find how far the avg might be from the
  mean with a certain probability

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Ch 8, Confidence

• This leads us to say that we can be, say, 95
  confident that the mean is within a certain
  distance of the avg
• Note that the mean might be unknown, but it is
  not random
• It does not change if we take a different sample
• So we cannot assign probabilities to the mean

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Ch 8, Confidence

• If we give a 95 confidence interval for the
  mean, we are 95 confident that the true mean
  lies within this interval
• The interval will be based on an interval that
  has a 95 probability of containing the avg
• (Since the avg IS random, we can assign
  probabilities to it.)

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Ch 8, Confidence

• Suppose we have a normal distn with SD11.5
• Consider the avg of N6 values from this
  population
• In what interval (around the mean) will the avg
  fall with probability 95?

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Ch 8, Confidence
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Ch 8, Confidence

• So 95 of the time, the avg will be within /-9.2
  of the mean
• We now reverse this and say that we are 95
  confident that the mean is within /-9.2 of the
  avg
• If the avg turns out to be 63.7, then the 95 CI
  for the mean is 63.7 /-9.2

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Ch 8, Confidence
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Ch 8, Confidence

• Procedure
• For given confidence, find how many SD the avg
  and mean might be apart
• We are confident that the mean is within this far
  of the avg

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Ch 8, Confidence

• Example
• We have a normal population with SD8.4
• The avg of a sample of size 7 is 33.5
• Find 90 CI for the mean of the population

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Ch 8, Confidence
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Ch 8, Confidence

• So the avg might be 1.6458.4/sqrt(7) away from
  the mean
• We are 90 confident that the mean is no higher
  than 33.5 1.6458.4/sqrt(7) and no lower than
  35- 1.6458.4/sqrt(7)

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Ch 8, Confidence

• In the previous problems, we had to assume we
  knew the value for the SD
• This is generally not true
• Generally we have to estimate SD from the data
• Replace the normal distn with Students t distn

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Ch 8, Confidence

• Students t is similar to normal, but has an
  extra parameter
• Degrees of freedom (df)
• Measures how good our est of SD is
• FOR THESE PROBLEMS, dfN-1
• For large df, t is essentially normal
• For smaller df, the quantiles are a little larger
  for t than for normal

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Ch 8, Confidence

• (See new PROBCALC.XLS)
• For normal, 90 CI is /-1.645 SD
• For t with 5 df, 90 is 2.015 SD
• For df15, 90 is 1.753 SD
• For df75, 90 is 1.665 SD

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Ch 8, Assumptions

• We use the t distn when we estimate the SD from
  the data
• We are still assuming that the data comes from a
  normal population
• There are two issues

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Ch 8, Assumptions

• There may be outliers in the sample
• This means that some of the observations should
  not be considered to be representative of the
  underlying population
• For our purposes, we will only consider an
  observation to be an outlier if it is WAY outside
  the others

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Ch 8, Assumptions

• Boxplots may be helpful in suggesting outliers
• These are marked with asterisks
• But simply because a value is far from the other
  values does not make it an outlier
• There should be some reason to think that this
  point should not have been in the sample

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Ch 8, Assumptions

• Suppose we take a sample of cities
• New York might be an outlier because it is so
  different from other cities
• If we consider the stock market, we might want to
  eliminate 9/11
• If we consider Olympic records, we might ignore
  Bob Beamons 1968 record

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Ch 8, Assumptions

• The other assumption is that the underlying distn
  is normal
• A good way to assess whether the distn is normal
  is to use a normal probability plot
• This plots the sorted data vs percentiles of the
  (standard) normal
• If the plot is approx a straight line, then the
  data appears to come from a normal distn

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Ch 8, Assumptions

• How close to a straight line?
• We will only be concerned if the plot is
  OBVIOUSLY not a straight line

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Ch 8, Assumptions

• What to do if the distn is NOT normal?
• This changes the question
• The mean is the obvious way to describe the
  normal distn
• For a different distn, it is less clear what to
  use to describe

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Ch 8, Assumptions

• We might consider using the median to describe
  the distn
• If we consider a possible value for the median
  and count the number in our sample that are above
  (or below) this value, we get the binomial distn
• Not so simple to form confidence intervals