Chapter 14 Tests of Hypotheses Based on Count Data

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Chapter 14 Tests of Hypotheses Based on Count
Data

• 14.2 Tests concerning proportions (large samples)
• 14.3 Differences between proportions
• 14.4 The analysis of an r x c table

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14.2 Tests concerning proportions (large samples)

• npgt5 n(1-p)gt5
• n independent trials
• X of successes
• pprobability of a success
• Estimate

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Tests of Hypotheses

• Null H0 pp0
• Possible Alternatives
• HA pltp0
• HA pgtp0
• HA p?p0

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Test Statistics

• Under H0, pp0, and
• Statistic
• is approximately standard normal under H0 .
• Reject H0 if z is too far from 0 in either
  direction.

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Rejection Regions
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Equivalent Form
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Example 14.1

• H0 p0.75 vs HA p?0.75
• ?0.05
• n300
• x206
• Reject H0 if zlt-1.96 or zgt1.96

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Observed z value

• Conclusion reject H0 since zlt-1.96
• P(zlt-2.5 or zgt2.5)0.0124lta ?reject H0.

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Example 14.2

• Toss a coin 100 times and you get 45 heads
• Estimate pprobability of getting a head
• Is the coin balanced one? a0.05
• Solution
• H0 p0.50 vs HA p?0.50

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Enough Evidence to Reject H0?

• Critical value z0.0251.96
• Reject H0 if zgt1.96 or zlt-1.96
• Conclusion accept H0

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Another example

• The following table is for a certain screening
  test

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• Test to see if the sensitivity of the screening
  test is less than 97.
• Hypothesis
• Test statistic

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What is the conclusion?

• Check p-value when z-2.6325, p-value 0.004
• Conclusion we can reject the null hypothesis at
  level 0.05.

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One word of caution about sample size

• If we decrease the sample size by a factor of 10,

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And if we try to use the z-test,
P-value is greater than 0.05 for sure (p0.2026).
So we cannot reach the same conclusion.
And this is wrong!
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So for test concerning proportions

• We want
• npgt5 n(1-p)gt5

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14.3 Differences Between Proportions

• Two drugs (two treatments)
• p1 percentage of patients recovered after taking
  drug 1
• p2 percentage of patients recovered after taking
  drug 2
• Compare effectiveness of two drugs

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Tests of Hypotheses

• Null H0 p1p2 (p1-p2 0)
• Possible Alternatives
• HA p1ltp2
• HA p1gtp2
• HA p1?p2

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Compare Two Proportions

• Drug 1 n1 patients, x1 recovered
• Drug 2 n2 patients, x2 recovered
• Estimates
• Statistic for test
• If we did this study over and over and drew a
  histogram of the resulting values of ,
  that histogram or distribution would have
  standard deviation

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Estimating the Standard Error

• Under H0, p1p2p. So
• Estimate the common p by

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So put them together
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Example 12.3

• Two sided test
• H0 p1p2 vs HA p1?p2
• n180, x156
• n280, x238

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Two Tailed Test

• Observed z-value
• Critical value for two-tailed test 1.96
• Conclusion Reject H0 since zgt1.96

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Rejection Regions
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P-value of the previous example

• P-valueP(zlt-2.88)P(zgt2.88)20.004
• So not only we can reject H0 at 0.05 level, we
  can also reject at 0.01 level.

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14.4 The analysis of an r x c table

• Recall Example 12.3
• Two sided test H0 p1p2 vs HA
  p1?p2
• n180, x156 n280, x238
• We can put this into a 2x2 table and the question
  now becomes is there a relationship between
  treatment and outcome? We will come back to this
  example after we introduce 2x2 tables and
  chi-square test.

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2x2 Contingency Table

• The table shows the data from a study of 91
  patients who had a myocardial infarction (Snow
  1965). One variable is treatment (propranolol
  versus a placebo), and the other is outcome
  (survival for at least 28 days versus death
  within 28 days).

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Hypotheses for Two-way Tables

• The hypotheses for two-way tables are very broad
  stroke.
• The null hypothesis H0 is simply that there is no
  association between the row and column variable.

• The alternative hypothesis Ha is that there is an
  association between the two variables. It
  doesnt specify a particular direction and cant
  really be described as one-sided or two-sided.

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Hypothesis statement in Our Example

• Null hypothesis the method of treating the
  myocardial infarction patients did not influence
  the proportion of patients who survived for at
  least 28 days.
• The alternative hypothesis is that the outcome
  (survival or death) depended on the treatment,
  meaning that the outcomes was the dependent
  variable and the treatment was the independent
  variable.

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Calculation of Expected Cell Count

• To test the null hypothesis, we compare the
  observed cell counts (or frequencies) to the
  expected cell counts (also called the expected
  frequencies)
• The process of comparing the observed counts with
  the expected counts is called a goodness-of-fit
  test. (If the chi-square value is small, the fit
  is good and the null hypothesis is not rejected.)

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• Observed cell counts

Expected cell counts
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The Chi-Square ( c2) Test Statistic
The chi-square statistic is a measure of how much
the observed cell counts in a two-way table
differ from the expected cell counts. It can be
used for tables larger than 2 x 2, if the average
of the expected cell counts is gt 5 and the
smallest expected cell count is gt 1 and for 2 x
2 tables when all 4 expected cell counts are gt 5.
The formula is c2 S(observed count expected
count)2/expected count Degrees of freedom (df)
(r 1) x (c 1) Where observed is an observed
sample count and expected is the computed
expected cell count for the same cell, r is the
number of rows, c is the number of columns, and
the sum (S) is over all the r x c cells in the
table (these do not include the total cells).
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The Chi-Square ( c2) Test Statistic
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Example Patient Compliance w/ Rx
In a study of 100 patients with hypertension, 50
were randomly allocated to a group prescribed 10
mg lisinopril to be taken once daily, while the
other 50 patients were prescribed 5 mg lisinopril
to be taken twice daily. At the end of the 60
day study period the patients returned their
remaining medication to the research pharmacy.
The pharmacy then counted the remaining pills and
classified each patient as lt 95 or 95
compliant with their prescription. The two-way
table for Compliance and Treatment was
Treatment Compliance 10 mg Daily 5 mg
bid Total 95 46 40 86 lt 95 4 10 14 Total
50 50 100
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Example Patient Compliance w/ Rx
Treatment Compliance 10 mg Daily 5 mg
bid Total 95 460 400 860 lt 95
40 100 140 Total 500 500 1000
c2 29.9, df (2-1)(2-1) 1, P-value lt0.001
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If we use the two sample test for proportion
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The c2 and z Test Statistics
The comparison of the proportions of successes
in two populations leads to a 2 x 2 table, so the
population proportions can be compared either
using the c2 test or the two-sample z test . It
really doesnt matter, because they always give
exactly the same result, because the c2 is equal
to the square of the z statistic and the
chi-square with one degree of freedom c2(1)
critical values are equal to the squares of the
corresponding z critical values.

• A P-value for the 2 x 2 c2 can be found by
  calculating the square root of the chi-square,
  looking that up in Table for P(Z gt z) and
  multiplying by 2, because the chi-square always
  tests the two-sided alternative.

• For a 2 x 2 table with a one-sided alternative
  hypothesis the two-sample z statistic would need
  to be used.

• To test more than two populations the chi-square
  must be used

• The chi-square is the one most often seen in the
  literature

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Summary Computations for Two-way Tables

• create the table, including observed cell counts,
  column and row totals.

• Find the expected cell counts.
• Determine if a c2 test is appropriate
• Calculate the c2 statistic and number of degrees
  of freedom

• Find the approximate P-value
• use Table III chi-square table to find the
  approximate P-value
• or use z-table and find the two-tailed p-value if
  it is 2 x 2.

• Draw conclusions about the association between
  the row and column variables.

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Yates Correction for Continuity

• The chi-square test is based on the normal
  approximation of the binomial distribution
  (discrete), many statisticians believe a
  correction for continuity is needed.
• It makes little difference if the numbers in the
  table are large, but in tables with small numbers
  it is worth doing.
• It reduces the size of the chi-square value and
  so reduces the chance of finding a statistically
  significant difference, so that correction for
  continuity makes the test more conservative.

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What do we do if the expected values in any of
the cells in a 2x2 table is below 5?
For example, a sample of teenagers might be
divided into male and female on the one hand, and
those that are and are not currently dieting on
the other. We hypothesize, perhaps, that the
proportion of dieting individuals is higher among
the women than among the men, and we want to test
whether any difference of proportions that we
observe is significant. The data might look like
this

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The question we ask about these data is knowing
that 10 of these 24 teenagers are dieters, what
is the probability that these 10 dieters would be
so unevenly distributed between the girls and the
boys? If we were to choose 10 of the teenagers at
random, what is the probability that 9 of them
would be among the 12 girls, and only 1 from
among the 12 boys? --Hypergeometric
distribution! --Fishers exact test uses
hypergeometric distribution to calculate the
exact probability of obtaining such set of the
values.
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Fishers exact test

• Before we proceed with the Fisher test, we first
  introduce some notation. We represent the cells
  by the letters a, b, c and d, call the totals
  across rows and columns marginal totals, and
  represent the grand total by n. So the table now
  looks like this

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Fisher showed that the probability of obtaining
any such set of values was given by the
hypergeometric distribution                    
                                               
   
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In our example
More extreme than observed
As extreme as observed
Recall that p-value is the probability of
observing data as extreme or more extreme if the
null hypothesis is true. So the p-value is this
problem is 0.00137.
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The fisher Exact Probability Test

• Used when one or more of the expected counts in a
  contingency table is small (lt2).
• Fisher's Exact Test is based on exact
  probabilities from a specific distribution (the
  hypergeometric distribution).
• There's really no lower bound on the amount of
  data that is needed for Fisher's Exact Test. You
  can use Fisher's Exact Test when one of the cells
  in your table has a zero in it. Fisher's Exact
  Test is also very useful for highly imbalanced
  tables. If one or two of the cells in a two by
  two table have numbers in the thousands and one
  or two of the other cells has numbers less than
  5, you can still use Fisher's Exact Test.
• Fisher's Exact Test has no formal test statistic
  and no critical value, and it only gives you a
  p-value.