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Hypothesis Test for a Population Proportion—Small Sample Case

 

Example:  According to the United States Department of Agriculture, 48.9%  of males between 20 and 39 years of age consume the minimum daily requirement of calcium.  After an aggressive “got milk” advertising campaign, the USDA conducts a survey of 35 randomly selected males between the ages of 20 and 39 and finds that 21 of them consume the recommended daily allowance of calcium.  At the  level of significance, is there evidence to conclude that the percentage of males between the ages of 20 and 39 who consume the recommended daily allowance of calcium has increased?

In this test of a population proportion, the requirement is not satisfied.  (The calculation 35*.489*(1-.489) is equal to 8.75.)   (Note:  In cases in which the sample size is relatively small this requirement is often not satisfied.)

 

An alternative method of testing a hypothesis about a population proportion is to use the binomial probability formula to calculate the likelihood of the sample result.  If the sample result is unusual then we will reject the null hypothesis.  We define unusual events as events that have a probability less than .05.

 

The hypothesis test is:    vs.  .  The sample statistics are n=35 and X=21.  Using the binomial probability formula, we calculate the likelihood of obtaining 21 or more males who consume the recommended daily allowance of calcium is .489. 

 

Press 1 -2nd DISTR and select A:binomcdf(.  Type in 35 , .489 , 20 ).

 

 

(Note:  The command binomcdf calculates the probability that , which is the complement of the probability that   To obtain , we calculate  and subtract this value from 1.)

 

The result is .  Since this probability is greater than .05, the correct conclusion is to Fail to Reject .