The web based company Oh Baby! has a goal of processing 95% of its orders on the same day they are received. If 485 out of the next 500 orders are processed on the same day, would this prove that they are exceeding their goal, using a = .025?

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  1. Null hypothesis:
    Ho: p = .95 -->meaning: population proportion is equal to .95
    Alternative hypothesis:
    Ha: p > .95 -->meaning: population proportion is greater than .95

    Using a formula for a binomial proportion one-sample z-test with your data included, we have:
    z = .97 - .95 -->test value (485/500 = .97) minus population value (.95) divided by
    √[(.95)(.05)/500] -->.05 is (1 - .95) and 500 is the sample size.

    Finish the calculation.

    Using a z-table, find your critical or cutoff value using .025 level of significance for a one-tailed test. The test is one-tailed because the alternative hypothesis is showing a specific direction (greater than).

    If the test statistic exceeds the critical or cutoff value from the table, the null is rejected in favor of the alternative hypothesis and p > .95 (there is enough evidence to prove they are exceeding their goal).

    Hope this will help get you started.

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  2. Thank you for your help.

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