Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 87, x = 26; 98 percent (0.185, 0.413) (0.202, 0.396) (0.184, 0.414) (0.203, 0.395)

Respuesta :

Answer:

(0.185, 0.413)

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 interval [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 = 87, x = 26, p = \frac{x}{n} = \frac{26}{87} = 0.2989[/tex]

98% confidence interval

So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.325[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2989 - 2.325\sqrt{\frac{0.2989*0.7011}{87}} = 0.185[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2989 + 2.325\sqrt{\frac{0.2989*0.7011}{87}} = 0.413[/tex]

So the correct answer is:

(0.185, 0.413)

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