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recent suggestion is 60 min of exercise per day. A sample of 50 adults in a study of risk factor report exercising a mean of 38 min per day with a standerd deviation of 19 min .based on sampal data, is physical activity significantly less than recommended? Run the appropriate test at a 5% level of significance.

  • biostatistic -

    Use a one-sample z-test.

    z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

    With your data:
    z = (38 - 60)/(19/√50) = ?

    Finish the calculation.

    Check a z-table at .05 level of significance for a one-tailed test.
    If the z-test statistic exceeds the critical value from the z-table, reject the null and conclude ยต < 60 (which is the alternative hypothesis). If the z-test statistic does not exceed the critical value from the z-table, do not reject the null.

    I hope this will help get you started.

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