do people who exercise regularly score higher than people in general on a measure of health? for people in general, the average on this test is 80 with a standard deviation of 12. the distribution is normal. a researcher tested a sample of 30 people who exercise and found them to have a mean score of 86. use the .05 significance lever for this problem

1. **Identify the null and alternative hypotheses:**

- Null hypothesis (\(H_0\)): People who exercise regularly have the same health score as the general population, so \(\mu = 80\).
- Alternative hypothesis (\(H_1\)): People who exercise regularly have a higher health score than the general population, so \(\mu > 80\).

2. **Calculate the z-score:**

- The formula for the z-score is
z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

- Where:
\(\bar{x} = 86\) (sample mean),
\(\mu = 80\) (population mean),
\(\sigma = 12\) (population standard deviation),
\(n = 30\) (sample size).

- Plugging in the values:
z = \frac{86 - 80}{\frac{12}{\sqrt{30}}}
z = \frac{6}{\frac{12}{\sqrt{30}}}
z = \frac{6}{2.19}
z = 2.74

3. **Determine the critical z-value:**

- At the .05 significance level, the critical z-value for a one-tailed test can be found using a standard normal distribution table, which is approximately 1.645.

4. **Compare the calculated z-score to the critical z-value:**

- Since the calculated z-score \(z = 2.74\) is greater than the critical z-value \(z = 1.645\), we reject the null hypothesis.

5. **Conclusion:**

- People who exercise regularly score higher on the measure of health than people in general at the 0.05 significance level.