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Thread: hypothesis testing

  1. #1
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    hypothesis testing

    Hello all--I am having trouble with this question--I feel I understand the concept, but I keep getting a test (z) statistic above 12, which seems way too high. Help??

    The average number of employees in medium sized US businesses (population) is 110. A business administrator at a certain business claims that the average number is different from 110. A random of 40 business were chosen. The sample mean score was 130 and population standard deviation is 10. Is there sufficient evidence to support the business administratorís claim? Take alpha =.05 and use (+/-) 1.96 as the critical value.

    1 Determine the test statistic for the test. Round to the nearest hundredth.
    2 Should the null hypothesis be accepted or rejected?
    3 Is the test statistically significant? Provide your rationale for the conclusion.
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  2. #2
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    Re: hypothesis testing

    Quote Originally Posted by mathizfun View Post
    Hello all--I am having trouble with this question--I feel I understand the concept, but I keep getting a test (z) statistic above 12, which seems way too high. Help??

    The average number of employees in medium sized US businesses (population) is 110. A business administrator at a certain business claims that the average number is different from 110. A random of 40 business were chosen. The sample mean score was 130 and population standard deviation is 10. Is there sufficient evidence to support the business administrator’s claim? Take alpha =.05 and use (+/-) 1.96 as the critical value.

    1 Determine the test statistic for the test. Round to the nearest hundredth.
    2 Should the null hypothesis be accepted or rejected?
    3 Is the test statistically significant? Provide your rationale for the conclusion.
    H_0: \mu=110
    H_1: \mu\neq110

    Note that I would use H_1: \mu>110, but it appears your problem wants a two-tailed test.

    The test statistic is Z=\dfrac{\bar{x}-\mu_0}{\sigma} = \dfrac{130-110}{10} = 2.

    Here is why you got Z>12: You divided the standard deviation by the square root of the sample size. You should only divide the standard deviation by the square root of the sample size to get a standard error term for the t-test, which uses degrees of freedom. You can try this problem out using the t test and appropriate degrees of freedom and the test statistic T will be very similar to Z because the sample size is greater than 30. When N>30, the Z test is a sufficient approximation of the T-Test. Additionally, it is usually safe to assume \sigma = s for large sample sizes.
    Last edited by abender; Jul 15th 2017 at 11:21 PM.
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