SM239

1
SM239

• SM239 began with probability models
• Binomial, negative binomial
• Poisson, Erlang(?)
• Normal

2
SM239

• Statistics ideas
• Confidence bounds/ intervals for parameters
• Hypothesis testing
• Rejection regions, critical values, power(?)
• P-values after data has been collected

3
SM239

• Applied these for
• 1 mean
• 1 proportion

4
SM239

• This semester, we will cover
• 2 means, 2 proportions
• gt2 proportions
• gt2 means
• Linear regression yf(x1, x2, )
• Principal Components when the observations are
  vectors

5
SM239

• To find a 95 upper confidence bound for p in
  binomial, given x successes in N trials
• Find p to solve 0.05 Prob(x or fewer successes)
• To find lower bound, solve 0.05 Prob(x or more
  successes)

6
SM239

• To solve f(x)k, where f(x) is a continuous fn of
  x
• Bisect(_at_(x) f(x), k, low, hi, tol)
• The soln must lie in the range lowlt xlt hi
• Result will be accurate to tol

7
SM239

• For upper bound when we observe 4 successes in 11
  trials
• Bisect(_at_(p) sum(binopdf(04,11,p)), .05, .3, .9,
  .0001)

8
SM239

• To find 90 lower conf bound for mean when we
  observe avg and SD is known
• Solve for mean where 0.10 prob(avg or greater
  for given mean)

9
SM239

• Suppose avg23, SD5.1, N7
• Bisect(_at_(m) 1-normcdf(23,m, 5.1/sqrt(7)), .1, 10,
  23, .00001)
• Can also use shortcut method with norminv
• Draw the picture!

10
SM239

• Find critical value to test
• H0 mean17
• Ha meanlt17
• Based on avg when SD3.8, N5 and alpha0.01

11
SM239

• Using shortcut method
• Critnorminv(.01,17,3.8/sqrt(5))
• Using bisect
• Bisect(_at_(x) normcdf(x,17, 3.8/sqrt(5)), .01,
  10,17, .00001)

12
Power

• Power is the prob we reject H0 when H0 is false
• Since Ha is typically of the form meanlt25 or p
  not 0.5, we must specify the parameter value
  where we are computing power

13
Power

• Suppose we test
• H0 mean15
• Ha meanlt15
• Based on avg of 7 values when we assume SD6.3
• We let the RR be avglt12

14
Power

• Note that alpha is given by
• gtgt normcdf(12,15,6.3/sqrt(7))
• 0.1039
• To find the power when the mean is 13, we find
  the prob of RR when mean13
• gtgt normcdf(12,13,6.3/sqrt(7))
• 0.3373
• So when the mean is 12, there is a 34 chance
  that we will (correctly) reject H0

15
Power

• The power when mean10 is
• gtgt normcdf(12,10,6.3/sqrt(7))
• 0.7995
• From the picture, we can see that power increases
  as the mean decreases (in this case)

16
Power

• We can use alpha and power to determine sample
  size in the normal case
• (1) The critical value depends on alpha and N
• (2) For a given critical value, power depends on
  N
• Together, these can determine N

17
Power

• In the previous problem, suppose we set
  alpha0.10 and want the power to be 0.75 when
  mean13
• Guess and test
• gtgt n20
• gtgt critnorminv(.1, 15,6.3/sqrt(n))
• 13.1946
• gtgt powernormcdf(crit,13,6.3/sqrt(n))
• 0.5549
• Power too low. Need to increase N

18
Power

• gtgt n30
• gtgt critnorminv(.1, 15,6.3/sqrt(n))
• 13.5259
• gtgt powernormcdf(crit,13,6.3/sqrt(n))
• 0.6763
• Closer. Need to increase N more

19
Power

• gtgt bisect(_at_(n) normcdf( norminv(.1,
  15,6.3/sqrt(n)),13,6.3/sqrt(n)), .75,20,100,.1)
• ans
• 37.8906
• So, N38

20
Sign Test

• We always begin with a probability model and then
  work with parameters of the distribution
• P in binomial
• Mean in normal

21
Sign Test

• Nonparametrics
• Do not specify/ assume a probability model
• Methods that work regardless of the underlying
  distn

22
Sign Test

• The (population) Median has the property that
  Prob(XgtMedian) Prob(Xlt Median) ½
• (Assume a continuous distn)

23
Sign Test

• Suppose we wish to test
• H0 Median19
• Ha Medianlt19
• Based on a sample of 12 observations
• Our statistic will be the of observations that
  are lt19
• (Could also use that are gt19)

24
Sign Test

• Suppose we want alphalt0.10
• Reject H0 if Prob(Xgtk) lt 0.10
• And X has binomial distn with p1/2
• gtgt sum(binopdf(812,12,.5))
• 0.1938
• gtgt sum(binopdf(912,12,.5))
• 0.0730
• So reject H0 if 9 or more values are lt19

25
Sign Test

• Suppose that 10 values are lt19
• Find p-value
• gtgt sum(binopdf(1012,12,.5))
• 0.0193

26
Sign Test

• Can find confidence intervals by finding value of
  median so that tail probability is desired value
• Suppose that we have a sample of size 20 and want
  to find a 85 lower conf bound for the median
• Use values below median as our statistic
• A median is OK if Prob(statistic) is at least 15

27
Sign Test

• gtgt sum(binopdf(08,20,.5))
• 0.2517
• gtgt sum(binopdf(07,20,.5))
• 0.1316
• So, if there are 8 values below a given median,
  it is OK
• If there are only 7 below, then that median is
  NOT in confidence bound
• The 75 lower conf bound is (just above) the 8th
  smallest value

28
Paired Sign Test

• Paired data consists of two measurements on each
  subject
• Helpful when there might be differences between
  subjects that would confuse the results

29
Paired Sign Test

• Want to know if quiet meditation will improve
  math scores
• Give some math problems to students
• Then have students spend 5 minutes in quiet
  meditation
• Then give another test

30
Paired Sign Test

• We could see how many students did better on the
  second test than on the first
• Suppose that 16 of 25 students did better on the
  second test

31
Paired Sign Test

• H0 prob(better on 2nd test)1/2
• Ha probgt1/2 if meditation helps
• P-value prob(16 or more of 25 did better on 2nd
  test)
• gtgt sum(binopdf(1625,25,.5))
• 0.1148

32
Paired Sign Test

• See Table 8.15, p 341
• 10 subjects given 2 types of aspirin
• Want to know if there is any difference between
  the types of aspirin
• For 8 patients, Type A was higher
• For 1 patient (5), they were the same
• For 1 patient (6), B was higher

33
Paired Sign Test

• When there is a tie (both the same) we usually
  omit this from our data
• So, of the 9 where there is a difference, 8
  showed A to be higher
• gtgt sum(binopdf(89,9,.5))
• 0.0195
• But the question was two-sided, so p-value4
• This is small, so we conclude that there is a
  difference between A and B