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More on Test the Hypothesis about a Population Proportion

Example1: In a survey conducted by the American Animal Hospital Association, 37% of respondents stated that they talk to their pets on the answering machine or telephone.  A veterinarian found this result hard to believe, so he randomly selected 150 pet owners and discovered that 54 of them spoke to their pet on the answering machine or telephone.  Test the veterinarian’s claim that less than 37% of pet owners speak to their pets on the answering machine or telephone, at the  level of significance.

This hypothesis test is a left-tailed test of :  vs.    To use the 1-Proportion Test, first you must determine whether the requirements for this test have been satisfied.  The first requirement is:   To  verify this, calculate 150*.37*(1-.37).  The result is greater than 10, so the first requirement is satisfied.  The second requirement is that the sample size is not more than 5% of the population size.  In this example, the population is all pet owners.  We don’t know the exact size of the population, but it is in the millions.  The sample size of 150 is definitely less than 5% of the population size.

To run the test, press STAT, highlight TESTS and select 5:1-PropZTest.  Enter .37 for .  For X, enter 54 and for n, enter 150.  Select for the alternative hypothesis and press ENTER.

Highlight Calculate and press ENTER.

Or, highlight Draw and press ENTER.

Since the P-value is greater than a, the correct conclusion is to Fail to Reject  .  The veterinarian’s claim that less than 37% of pet owners speak to their pets is not supported by the data.

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Example 2:

Small-Sample Hypothesis Test   In 1997, 4% of mothers smoked more than 21 cigarettes during their pregnancy.  An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than 4% today.  She randomly selects 120 pregnant mothers and finds that 3 of them smoked 21 or more cigarettes during pregnancy.  Test the researcher’s claim at the level of significance

In this test of a population proportion, the requirement is not satisfied.  (The calculation 120*.04*(1-.04) is equal to 4.608.)  (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 that .05. 

The hypothesis test is:  vs. .   The sample statistics are n=120 and x=3.  Using the binomial probability formula, we calculate the likelihood of obtaining 3 or fewer mothers who smoked 21 or more cigarettes during pregnancy.  We assume that the proportion of mothers who smoked 21 or more cigarettes during pregnancy is .04. 

Press 2^nd DISTR and select A:binomcdf(.   Type in 120 , .04 , 3 ).

The result is:    Since this probability is greater than .05, the correct conclusion is to Fail to Reject .  The data does not support the obstetrician’s belief that less than 4% of mothers smoked 21 or more cigarettes during pregnancy.

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