Tests with One Sample, Dichotomous Outcome
Hypothesis testing applications with a dichotomous outcome variable in a single population are also performed according to the five-step procedure. Similar to tests for means, a key component is setting up the null and research hypotheses. The objective is to compare the proportion of successes in a single population to a known proportion (p0). That known proportion is generally derived from another study or report and is sometimes called a historical control. It is important in setting up the hypotheses in a one sample test that the proportion specified in the null hypothesis is a fair and reasonable comparator.
In one sample tests for a dichotomous outcome, we set up our hypotheses against an appropriate comparator. We select a sample and compute descriptive statistics on the sample data. Specifically, we compute the sample size (n) and the sample proportion which is computed by taking the ratio of the number of successes to the sample size,
We then determine the appropriate test statistic (Step 2) for the hypothesis test. The formula for the test statistic is given below.
Test Statistic for Testing H0: p = p 0
if min(np0 , n(1-p0))> 5
The formula above is appropriate for large samples, defined when the smaller of np0 and n(1-p0) is at least 5. This is similar, but not identical, to the condition required for appropriate use of the confidence interval formula for a population proportion, i.e.,
Here we use the proportion specified in the null hypothesis as the true proportion of successes rather than the sample proportion. If we fail to satisfy the condition, then alternative procedures, called exact methods must be used to test the hypothesis about the population proportion.
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Example:
The NCHS report indicated that in 2002 the prevalence of cigarette smoking among American adults was 21.1%. Data on prevalent smoking in n=3,536 participants who attended the seventh examination of the Offspring in the Framingham Heart Study indicated that 482/3,536 = 13.6% of the respondents were currently smoking at the time of the exam. Suppose we want to assess whether the prevalence of smoking is lower in the Framingham Offspring sample given the focus on cardiovascular health in that community. Is there evidence of a statistically lower prevalence of smoking in the Framingham Offspring study as compared to the prevalence among all Americans?
- Step 1. Set up hypotheses and determine level of significance
H0: p = 0.211 H1: p < 0.211 α=0.05
- Step 2. Select the appropriate test statistic.
We must first check that the sample size is adequate. Specifically, we need to check min(np0, n(1-p0)) = min( 3,536(0.211), 3,536(1-0.211))=min(746, 2790)=746. The sample size is more than adequate so the following formula can be used:
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- Step 3. Set up decision rule.
This is a lower tailed test, using a Z statistic and a 5% level of significance. Reject H0 if Z < -1.645.
- Step 4. Compute the test statistic.
We now substitute the sample data into the formula for the test statistic identified in Step 2.
- Step 5. Conclusion.
We reject H0 because -10.93 < -1.645. We have statistically significant evidence at α=0.05 to show that the prevalence of smoking in the Framingham Offspring is lower than the prevalence nationally (21.1%). Here, p < 0.0001.
The NCHS report indicated that in 2002, 75% of children aged 2 to 17 saw a dentist in the past year. An investigator wants to assess whether use of dental services is similar in children living in the city of Boston. A sample of 125 children aged 2 to 17 living in Boston are surveyed and 64 reported seeing a dentist over the past 12 months. Is there a significant difference in use of dental services between children living in Boston and the national data?
Calculate this on your own before checking the answer.
Video - Hypothesis Test for One Sample and a Dichotomous Outcome (3:55)
Link to transcript of the video
