Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight-year period.

Required:
a. Construct a 95% confidence interval for the population proportion of people over 50 who ran and died in the same eight-year period.
b. Explain in a complete sentence what the confidence interval means

Respuesta :

Answer:

a. The 95% confidence interval for the population proportion of people over 50 who ran and died in the same eight-year period is (0.0038, 0.0262).

b. It means that we are 95% sure that the true proportion of people over 50 who ran and died in the same eight-year period is (0.0038, 0.0262).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 451, \pi = 0.015[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.015 - 1.96\sqrt{\frac{0.015*0.985}{451}} = 0.0038[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.015 + 1.96\sqrt{\frac{0.015*0.985}{451}} = 0.0262[/tex]

The 95% confidence interval for the population proportion of people over 50 who ran and died in the same eight-year period is (0.0038, 0.0262).

b. Explain in a complete sentence what the confidence interval means

It means that we are 95% sure that the true proportion of people over 50 who ran and died in the same eight-year period is (0.0038, 0.0262).

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