## Medical researchers have devised a test that quantitatively meaasures a person’s risk of heart disease. The test scores for the general publ

Medical researchers have devised a test that quantitatively meaasures a person’s risk of heart disease. The test scores for the general public are approximately normally distributed with a mean of 50. On the test, higher test scores indicate a higher risk of heart disease. An advocate of long distance running believes that marathon runners should have a lower risk of heart disease than the general public. A random sample of 20 marathon runners is selected and each is tested for their risk of heart disease. Their test scores have a mean of 43.7 and a standard deviation of 4.2. Construct a 90% confidence interval for u, the true mean score for all marathon runners.

a. 43.7 + 1.62.

b. 43.7 + 6.91.

c. 43.7 + 0.94.

d. 43.7 + 4.2.

e. 43.7 + 8.40.

## Answers ( )

Answer:The confidence level for this case is 90% so then the significance level is and so then the critical value in the t distribution with df =19 is given by :

Now we can find the margin of error for the confidence interval given by:

And replacing we got:

And the confidence interval would be given by:

And the best option is:

a. 43.7 + 1.62.

Step-by-step explanation:The random variable of interest for this problem is the score for all marathon runners X and in special we want to infer about the true mean . We have the following info given:

the sample mean for the test scores

the sample deviation given

the sample size selected

And we want to construct a confidence interval for the true mean given by this formula:

The degrees of freedom for this case are given by:

The confidence level for this case is 90% so then the significance level is and so then the critical value in the t distribution with df =19 is given by :

Now we can find the margin of error for the confidence interval given by:

And replacing we got:

And the confidence interval would be given by:

And the best option is:

a. 43.7 + 1.62.