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March 29, 2017

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From a random sample of 500 males interviewed, 125 indicated that they watch professional football on Monday night television. Does this evidence indicate that more than 20% of the male TV viewers watch professional football on Monday evenings? Use the 0.01 level of significance, and find the p-value.

  • statistics - ,

    You can try a proportional one-sample z-test for this one since this problem is using proportions.

    Here's a few hints to get you started:

    Null hypothesis:
    Ho: p = .20 -->meaning: population proportion is equal to .20

    Alternative hypothesis:
    Ha: p > .20 -->meaning: population proportion is greater than .20

    Using a formula for a proportional one-sample z-test with your data included, we have:
    z = .25 - .20 -->test value (125/500 is .25) minus population value (.20) divided by
    √[(.25)(.75)/500] --> .75 represents 1-.25 and 500 is sample size.

    Finish the calculation. Remember if the null is not rejected, then you cannot conclude p > .20. If you need to find the p-value for the test statistic, check a z-table, then compare to .01 to determine whether or not to reject the null.

    I hope this helps.

  • Correction - ,

    This section should read:

    Using a formula for a proportional one-sample z-test with your data included, we have:
    z = .25 - .20 -->test value (125/500 is .25) minus population value (.20) divided by
    √[(.20)(.80)/500] --> .80 represents 1-.20 and 500 is sample size.

    Sorry for any confusion.

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