Posted by **nono** on Tuesday, November 20, 2012 at 1:37pm.

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 -
**MathGuru**, Tuesday, November 20, 2012 at 7:19pm
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.

## Answer This Question

## Related Questions

- accountung - a fixed mo fee of 22.00 up to 510 min. then .05 per min thereafter ...
- accounting - paying a fixed monthly fee of $15.00 up yo 240 min then .08 per min...
- Math - Please help me solve these problem i have no idea how to do this. 720m/d...
- Accounting - variable cost, fixed cost, total costif .10 per min long dist ...
- health - i was completing my study guide so i can study and i ran across these ...
- college-Stats - A certain geyser was determined to have eruptions with a Mean ...
- MATH!! - TO be on the 1-km race team, Celia must have a mean time less than 5 ...
- math - amos is using a computer program to graph the function N(p)=-73p-2818. He...
- math - 289 min = 4 hr and 49 min 213 min = 3 hr and 33 min 345 min = 5 hr and 45...
- statistics - Mean time 79 min Standard deviation 8 min what is the probability ...

More Related Questions