# 7.4.5 Single Sample Hypothesis Test for a Proportion

Here is the screenshot of the basic StatCrunch solution.

[My Excel calculator for running this test is found here.] As always for proportion problems, we have to check first to be sure np and nq or n*(1-p) are both > 5.

If they are, we can use the normal approximation to the binomial (a proportion is essentially a binomial test). That is why for all the proportion problems we do in this course we use the z-test. If either np or nq is not > 5, we have to use a test that is beyond the scope of this course.

Step 1 as almost always is true for us, is to click on Stat once StatCrunch is open. The sequence is Stat>Proportion Stats>One Sample>With Summary. Then enter the number of success, which is just n*p, the sample size, select the Hypothesis test for p, and enter the null hypothesis and select the math operator for Ha. Click Compute! The answer window opens and you can see the test statistic of z= -1.042 and the p-value of 0.29. Although this problem asks us to find the critical value and make our decision based on that, the p-value always agrees and tells us we Fail to Reject the null since p>α.

I used the Stat>Calculators>Normal path to bring up the normal calculator. Because this is a two-tailed test (recall Ha has the ≠ math operator), I like to use the Between button and enter the confidence level c, which is 1-α or 0.9. The calculator shows the two critical values and rejection areas. Because the test statistic of -1.042 does not fall in either rejection area, we get the same decision of Fail to Reject the null. The final part is to draw a conclusion. Recall the alternative was the claim, so we say: Fail to Reject Ho. The data do NOT provide sufficient evidence to Support the claim.

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