1. Lecture 6 Comparing Proportions (II)
Outline of Today:
Testing Equality of Proportions
The Chi-squared Test
9/19/2018
SA3202, Lecture 6

2. Testing Equality of Proportions
Hypothesis Testing A basic hypothesis of interest is
H0: p1=p2 means H0: “C no effect on R”
means H0: “C and R are independent”
This test is also known as “Independence Test”. The alternative may be
one-sided “H1: p1<p2”
Or “H1: p1>p2”,
or two sided , “ H1: p1 not equal to p2”.
Let =p1-p2 be the population proportion difference. Its estimator is the sample proportion difference
~A N( , )
where the variance is
The null hypothesis H0: p1=p2 is equivalent o H0: =0. The latter may be tested using
Z= ~ AN(0, 1) under H0
9/19/2018
SA3202, Lecture 6

3. Testing Equality of the Proportions
Estimating the Common Variance Since the variance is unknown, we need to estimate it under H0. Let p denote the common value of p1 and p2 under H0 : p1=p2=p. Then
Var( ||H0)=p1(1-p1)/n1+p2(1-p2)/n2=(1/n1+1/n2) p(1-p).
The best estimator of p is the Pooled estimator
(X1+X2)/(n1+n2)
So the standard error is
s.e( |H0)=
9/19/2018
SA3202, Lecture 6

4. Testing Equality of the Proportions
The Z- Test Statistic is then constructed as:
Z= /s.e( | H0) ~AN(0,1) under H0
Testing Procedure: H0: p1=p2 is tested by comparing Z with the percentage points of N(0,1)
if against H1: p1<p2, reject H0 if Z is too small,
against H1: p1>p2, reject H0 if Z is too large,
against H1: p1 not equal p2, reject H0 if |Z| is too large
9/19/2018
SA3202, Lecture 6

5. Example
For the Vitamin C Data, consider testing H0: Vitamin C has no effect (p1=p2) against the alternative H1: it is effective in preventing cold (p1>p2).
Placebo Vitamin C
Cold
Not Cold
Total
Now
the sample proportion difference =.0991,
the pooled sample proportion =(X1+X2)/(n1+n2)=(31+17) /( )=.1720
s.e. ( |H0)=.0448
Z= /s.e.( |H0 )=.0991/.0448=2.19,
At 5% level, Reject H0: p1=p2 in favor H1: p1>p2 since Z> This means that Vitamin C seems to have some effect. This is consistent with the conclusion made previously using the CI procedure.
9/19/2018
SA3202, Lecture 6

6. The Chi-squared Test
The H0: p1=p2 can also be tested by a chi-squared test: compare the observed frequencies with the expected frequencies using the Pearson’s Goodness of test of the Wilk’s Likelihood ratio test.
Response Group Group 2
Positive X X2
Negative n1-X n2-X2
Total n n2
Under H0: p1=p2=p, the expected frequencies are
Response Group Group 2
Positive n1p n2p
Negative n1(1-p) n2(1-p)
Total n n2
9/19/2018
SA3202, Lecture 6

7. The Chi-squared Test
Replacing the p by its pooled sample proportion , we get the contingency table of the estimated expected frequencies.
Response Group Group 2
Positive
Negative
Total n n2
Thus we can constructed the Pearson’s Goodness of fit Test statistic or the Wilk’s Likelihood Ratio Test statistic
The d.f. =4-3=1, with 3 constraints:
9/19/2018
SA3202, Lecture 6

8. Example
For the Vitamin C Data, consider testing H0: Vitamin C has no effect (p1=p2)
against the alternative H1: it is effective.
Placebo Vitamin C
Cold
Not Cold
Total
Under H0, the pooled sample proportion =(X1+X2)/(n1+n2)=(31+17) /( )=.1720
Under H0, the estimated expected frequencies are
Cold e.g =139*.172
Not Cold
The Pearson’s Goodness of Fit Test statistic is
T=( )^2/24.1+( )^2/115.9+( )^2/24.1+( )^2/115.1=4.61
The 95% table value with 1 df= Therefore, H0 is rejected . That is the effect of Vitamin is significant at 5% level. Note that the Chi-squared Test is applicable only for two-sided alternatives.
9/19/2018
SA3202, Lecture 6