8.1
Inference Section for a Single Proportion
Recall: Population Proportion
Let p be the proportion of "successes" in a population. A random sample of size n is selected, and X is the count of successes in the sample. Suppose n is small relative to the population size, so that X can be regarded as a binomial random variable with
X = np
and
X = np(1 - p)
Recall: Population Proportion
We use the sample proportion p = ^ n the population proportion p.
X
as an estimator of
^ p is an unbiased estimator of p, with mean and SD:
p = p ^
and
p = ^
p (1 - p ) n
^ When n is large, p is approximately normal. Thus
z= ^ p- p p (1 - p )
n
is approximately standard normal.
CI for a Population Proportion
^ Since p is normal. ^ The standard error of p is
^ SE ( p) =
^ ^ p (1 - p) n
An approximate level C confidence interval for p:
^ ^ ^ p z SE ( p ) = p z
*
*
^ ^ p(1 - p) n
where P(Z z*) = (1 C)/2.
CI for a Population Proportion
The margin of error is
^ m = z SE ( p ) = z
*
*
^ ^ p(1 - p ) n
Use this interval when the successes and failures are both at least 15 Use Table A or last row of Table D to find z*.
Example:
A news program constructs a call-in poll about a proposed city ban on handguns. 2372 people call in to the show. Of these, 1921 oppose the ban. Construct a 95% confidence interval for the true proportion of people who oppose the ban. What are the possible problems with the study design?
Solution:
Note: Since p is a proportion, if you ever get an upper limit value of > 1 or lower <0 while calculating the CI, replace by 1 and 0 (respectively).
Choosing a Sample Size
If we want to estimate the proportion p within a specified margin of error m, the required sample size is (at least):
n=
^ ^ ( p) ( 1- p) ( z ) m
2
* 2
Choosing a Sample Size
^ Since p is unknown before the data is collected, we use any prior information we have to get a rough known estimate, p*. 2
( p ) ( 1- p ) ( z ) n=
* * *
m2
This is especially important if you believe p is close to 0 or 1. Where might we find previous information about p?
If you have no information, we may replace p*, above, with 0.5 to obtain the most conservative * 2 * 2 sample size. ( 0.5) ( 1 - 0.5 ) ( z ) ( z )
n= m
2
=
4m 2
Example (handguns revisited):
Assume that we plan to ask randomly chosen people from the phone book. We would like to have a margin of error of 0.03=3%. How big a sample size should we have now?
Another example:
Suppose that the results of a survey of 2,000 television viewers at 11:40p.m. on Monday September 28, 1998 were recorded, and it was determined that 226 viewers watched "The Tonight Show." Estimate with 95% confidence the number of TVs tuned to "The Tonight Show" if there are 100 million potential television sets.
Testing for a single population proportion
^ When n is large, p is approximately normal, so
z=
^ p - p0 p0 (1 - p0 ) n
is approximately standard normal.
We may test H0: p = p0 against one of these: Ha: p > p0 Ha: p < p0 Ha: p p0
Large-sample Significance Test for a Population Proportion
The null hypothesis, H0: p = p0 The test statistic is
z=
^ p - p0 p0 (1 - p0 ) n
P-value P(Z z) P(Z z) 2P(Z | z |)
Alternative Hypothesis Ha: p > p0 Ha: p < p0 Ha: p p0
Large-sample Significance Test for a Population Proportion
How big does the sample size need to be? The general rule of thumb to use here, as before for approximation of binomial distribution by normal distribution, is
np0 10, n(1 - p0 ) 10
Example:
A claim is made that only 34% of all college students have part-time jobs. You are a little skeptical of this result and decide to conduct an experiment to show more students work. You get a sample of 100 college students and find that 47 of these students have parttime jobs. Conduct hypothesis a test with = 0.05 to determine whether more than 34% of college students have parttime jobs.
SAS Programs
proportion.doc
But mostly hand computations.
Section 8.2
Comparing Two Proportions
Comparing Two Proportions
Before we begin... Intuitively, how do you think we will be comparing two proportions?
Think in terms of two means, what did we do there?
Comparing Two Proportions
Notation:
Populatio Population Sample Count of n proportion size successes 1 2 p1 p2 n1 n2 X1 X2
Comparing Two Proportions
SRS of size n1 from a large population having proportion p1 of successes and an independent SRS of size n2 from another large population having proportion p2 of successes. ^ p1 is an estimator of p1
^ p2
is an estimator of p2: X
^ p1 =
1
n1
X2 ^ p2 = n2
Comparing Two Proportions: properties of estimators
We have
p1 = p1 ^ p2 = p2 ^
p1 = ^ p2 = ^
p1 (1 - p1 ) n1 p2 (1 - p2 ) n2
Comparing Two Proportions: approximate normality
Compare p1 and p2 means to examine p1 p2. This can be done by studying the difference ^ ^ p1 - p2 We obtain an approximate standard normal variable ^ ^ ( p1 - p2 ) - ( p1 - p2 ) z= p1 (1 - p1 ) p2 (1 - p2 ) + n1 n2
Significance Test Comparing Two Population Proportions
When p1 and p2 are unknown, we want to carry out hypothesis testing for
H0: p1 = p2 (same as p1 p2=0)
against one of the following alternatives: Ha: p1 > p2
Ha: p1 < p2 Ha: p1 p2
Comparing Two Population Proportions: Significance Test
Under the null hypothesis H0: p1 = p2, we view all the data as coming from a single population with proportion p1=p2=p (p unknown). The z-statistic becomes ^ ^ ^ ^ ( p1 - p2 ) ( p1 - p2 ) z= = p (1 - p ) p (1 - p ) 1 1 + p(1 - p ) + n n n1 n2 2 1 Since p is unknown, we use the pooled sample ^ proportion p to estimate it: p = X 1 + X 2 ^ n1 + n2
Comparing Two Population Proportions- Significance Test
Null hypothesis: H0: p1 = p2 ^ ^ ( p1 - p2 ) The test statistic: z = 1 1 ^ ^ p (1 - p ) + n1 n2 Alternative Hypothesis Ha: p1 > p2 Ha: p1 < p2 Ha: p1 p2 P-value P(Z z) P(Z z) 2*P(Z | z |)
Example:
In a highly-publicized study, doctors confirmed earlier observations that aspirin seems to help prevent heart attacks. The research project employed 21,996 male American physicians. Half of these took an aspirin tablet every other day, while the other half took a placebo on the same schedule. After 3 years, researchers determined that 139 of those who took aspirin and 239 of those who took placebo had had heart attacks. Determine whether these results indicate that aspirin is effective in reducing the incidence of heart attacks at significance level 0.05.
Confidence Interval:
We are interested in constructing a confidence interval for p1 p2. The estimate for this is ^ ^ p1 - p2 ^ ^ The standard error of p1 - p2 is defined as
SE p1 - p2 = ^ ^
^ ^ ^ ^ p1 (1 - p1 ) p2 (1 - p2 ) + n1 n2
Confidence Interval - Formula:
An approximate level C confidence interval for p1 p2 is given by *
^ ^ ^ ^ ( p1 - p2 ) z SE ( p1 - p2 )
*
^ ^ = ( p1 - p2 ) z
^ ^ ^ ^ p1 (1 - p1 ) p2 (1 - p2 ) + n1 n2
where P(Z z*) = (1 C)/2
Confidence Interval-margin of error:
The margin of error is given by
^ ^ m = z SE ( p1 - p2 ) = z
*
*
^ ^ ^ ^ p1 (1 - p1 ) p2 (1 - p2 ) + n1 n2
Use this interval when the number of successes and the number of failures in each sample are at least 10.
Example (Aspirin and Heart Attacks):
Estimate with 95% confidence the difference in proportion of men risking a heart attack (in the next 3 years) among aspirin takers and non-takers.
Relative risk:
^ p1 RR = ^ p2
Example: Calculate the relative risk for the aspirin example: Software gives confidence intervals (based on data) for population relative risk: p1 / p2 This is another way of comparing the two population proportions.