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

Question

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.

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Everleigh 3 weeks 2021-10-01T14:57:14+00:00 1 Answer 0

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    2021-10-01T14:58:46+00:00

    Answer:

     df = n-1=20-1=19

    The confidence level for this case is 90% so then the significance level is \alpha=0.1 and \alpha/2 =0.05 so then the critical value in the t distribution with df =19 is given by :

     t_{\alpha/2}= 1.729

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

     ME = t_{\alpha/2}\frac{s}{\sqrt{n}}

    And replacing we got:

     ME = 1.729 *\frac{4.2}{\sqrt{20}}= 1.624

    And the confidence interval would be given by:

     43.7 \pm 1.62

    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 \mu. We have the following info given:

    \bar X= 43.7 the sample mean for the test scores

     s= 4.2 the sample deviation given

    n = 20 the sample size selected

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

    \bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

    The degrees of freedom for this case are given by:

     df = n-1=20-1=19

    The confidence level for this case is 90% so then the significance level is \alpha=0.1 and \alpha/2 =0.05 so then the critical value in the t distribution with df =19 is given by :

     t_{\alpha/2}= 1.729

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

     ME = t_{\alpha/2}\frac{s}{\sqrt{n}}

    And replacing we got:

     ME = 1.729 *\frac{4.2}{\sqrt{20}}= 1.624

    And the confidence interval would be given by:

     43.7 \pm 1.62

    And the best option is:

    a. 43.7 + 1.62.

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45:7+7-4:2-5:5*4+35:2 =? ( )