SM219 Normal

1
Ch 7, Sampling

• Since a sample is only part of a population, the
  avg of the sample will depend on which part we
  look at
• See cars-by-loc.xls
• Take samples of 5 cars and average
• Diff samples have diff avgs
• None are exactly the avg of all the cars

2
Ch 7, Sampling

• Want to describe the prob distn of the avg of a
  sample
• Regardless of what the underlying distn is,
  averages are approximately normal
• If the underlying data has a normal distn, then
  avgs are exactly normal
• The larger the sample, the closer the distn is to
  normal

3
Ch 7, Sampling

• Normal distns are described by their mean and SD
• The mean of the distn of the avg is the same as
  the mean of the population
• The SD of the avg is the SD of the population /
  ?N

4
Ch 7, Sampling

• Example
• Suppose that the weight of mids has a normal
  distn with mean160 and SD8.5
• If we take the avg of a random sample of 6 mids,
  then the mean of this avg 160 and the SD8.5/
  ?6

5
Ch 7, Sampling

• P(a mid 170) 0.12
• P(avg of 6 mids 170) .002
• The avg is much more likely to be close to the
  mean (160) than is any single mid

6
Ch 7, Sampling

• If we increase the size of the sample, we
  decrease SD
• Which makes it even more likely for the avg to be
  close to the mean
• P(avg of 6 mids 170) .002
• P(avg of 10 mids 170) .0001

7
Ch 7, Sample size

• We can use this to determine how large N has to
  be
• Suppose we have a population with SD8.3
• We are going to average N random elements
• What does N have to be so that we can be 95 sure
  that the average is within 2 units of the (true)
  mean

8
Ch 7, Sample size

• Note that we do not need to know the true mean to
  determine N
• 95 of the time, the average will be within 1.96
  SD/vN of the true mean
• Solve 2 1.968.3/ vN
• N66.162
• We could use 66 or 67

9
Ch 7, Sample size

• Exercises
• 1. In previous, find N so that we are 90 sure
  that the avg will be with 0.6 units of the true
  mean
• 2. Repeat 1 if SD15.8
• 3. Repeat 2 to be 99 sure that avg is within
  0.25 units of true mean

10
Ch 7, Sample fraction

• Suppose we take a sample of mids and record 1 if
  left handed and 0 if not
• This is called a Bernoulli RV
• Call it X
• If pprob a random mid is left handed, not hard
  to show
• Mean(X)p
• SD(X)?p(1-p)

11
Ch 7, Sample fraction

• The average of X is the fraction of mids in our
  sample who are left handed
• Mean(fraction)p
• SD(frac) ?p(1-p) / ?N

12
Ch 7, Sample fraction

• Although X is clearly NOT normal, by CLT, the avg
  is approx normal
• We can then use the techniques for avgs to work
  with fractions

13
Ch 7, Sample fraction

• Example
• Suppose that 10 of mids are left handed
• If N250, I can be 90 sure that the fraction of
  LH in my sample will be AT MOST how much?

14
Ch 7, Sample fraction
15
Ch 7, Sample fraction

• Repeat if N500
• Ans0.1172
• As N increases, the fraction gets closer to p

16
Ch 7, Sample fraction

• Find N so that we are 90 sure that the fraction
  is within 5 of p
• Could do guess and test
• Could find how many SD corresponds to 90
• 1.645 SD

17
Ch 7, Sample fraction

• Solve 1.645 SD.05
• 1.645 ?p(1-p)/N 0.05
• With p0.1, N97.41
• (Prob round up)
• If we dont know p, we can use p0.5 because
  thats the most variable case
• Then N270.6