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a_sh-v [17]
2 years ago
12

A recent survey showed that among 100 randomly selected college seniors, 20 plan to attend graduate school and 80 do not. determ

ine a 95 % confidence interval for the population proportion of college seniors who plan to attend graduate school. (enter each answer rounded to three decimal places.)
Mathematics
1 answer:
Temka [501]2 years ago
8 0
The confidence interval would be (0.122, 0.278).

We first find our z-score.  We want a 95% confidence interval:

0.95/2 = 0.475

Looking this up in the z-table, (http://www.statisticshowto.com/tables/z-table/) we see the z-score is 1.96.

The formula we will use is:

p \pm z\sqrt{\frac{p(1-p)}{n}}

In this problem, p = 20/100 = 0.2, and n=100:

0.2 \pm 1.96\sqrt{\frac{(0.2)(0.8)}{100}}
\\
\\0.2\pm 0.0784
\\
\\(0.2-0.0784, 0.2+0.0784)
\\
\\(0.1216,0.2784)
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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