# Hypothesis Testing with One Sample: Determining Outcome Variable

Recent recommendations suggest 60 minutes of physical activity per day. A sample of 50 adults in a study of cardiovascular risk factors report exercising a mean of 38 minutes per day with a standard deviation of 19 minutes. Based on the sample data, is the physical activity significantly less than recommended? Run the appropriate test at a 5% level of significance.

What is the effect of the choice of alpha on the results of the study? For what type of study would you choose the alpha level of .01? of .05? of.10?

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#### Solution Preview

Hello,

Firstly let us go over what is the formula for significance testing

t = observed -expected / standard deviation / square root of the sample size

Our null hypothesis would be:

Ho: Students get the recommended daily activity of 60 min

The alternative would be:

Ha: Students get LESS then the recommended daily activity of 60 min

So if you had an entire list of numbers, excel would be able to calculate the t-test score for you. BUT since we are given the means and the standard deviations without any other info (the raw data), there is not really a single stats program or macro that we can employ. I used a series of formulas to get the info that we need.

But, using excel, we see that our t-test yields a test statistic of -4.093776102

Now, we can get our critical values to compare our test stat to.

At an alpha of 0.05, we get a critical value of -1.677

Since ...

#### Solution Summary

This solution assists with the hypothesis testing question.