Posted by nono on .
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.
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 µ < 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.