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Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book. From a random sample of 50 students, she found that 32 students read at least 1 book last month. Assuming all conditions for inference are met, which of the following defines a 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month?
a. 32 + or - 1.645 underroot(32)(18)/50.
b. 32 + or - + 1.96 underroot(32)(18)/50.
c. 0.64 + or - 1.282 underroot(0.64)(0.36)/50.
d. 0.64 + or - 1.645 underroot(0.64)(0.36)/50.
e. 0.64 + or - 1.96 underroot(0.64)(0.36)/50.

Respuesta :

Answer:

[tex]0.64 \pm 1.645\sqrt{\frac{0.64*0.36}{50}}[/tex], option D

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].

From a random sample of 50 students, she found that 32 students read at least 1 book last month.

This means that [tex]n = 50, \pi = \frac{32}{50} = 0.64[/tex]

90% confidence level

So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].

Confidence interval:

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

[tex]0.64 \pm 1.645\sqrt{\frac{0.64*0.36}{50}}[/tex]

So the correct answer is given by option D.

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