Statistics

A researcher believes that the percentage of people who exercise in California is greater than the national exercise rate. The national rate is 20%. The researcher gathers a random sample of 120 individuals who live in California and finds that the number who exercise regularly is 31 out of 120.

a. What is Zobt?

b. What is df for this test?

c. What is Chi-sqare cv?

d. What conclusion should be drawn from these results?

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  1. If you do a proportional z-test with this data, it would look something like this:

    z = (.258 - .20)/√[(.20)(.80)/120]

    Note: 31/120 = .258

    Calculating:
    z = .058/.0365 = 1.589

    The null would not be rejected and the conclusion would be that there is no difference in the population of interest.

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