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Volgvan
3 years ago
5

A random sample of 432 voters revealed that 100 are in favor of a certain bond issue. A 95 percent confidence interval for the p

roportion of the population of voters who are in favor of the bond issue is A 100 + 1.96 0.5(0.5) 432 100 + 1.645 0.5(0.5) 432 100 + 1.96 0.231(0.769) 432 0.231 +1.96 0.231(0.769) 432 0.231(0.709) 0.231 +1.6451 432​
Mathematics
1 answer:
kolbaska11 [484]3 years ago
4 0

Answer:

The 95% confidence interval is 0.231 \pm 1.96\sqrt{\frac{0.231*0.769}{432}}

Step-by-step explanation:

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

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

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

A random sample of 432 voters revealed that 100 are in favor of a certain bond issue.

This means that n = 432, \pi = \frac{100}{432} = 0.231

95% confidence level

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

Confidence interval:

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

0.231 \pm 1.96\sqrt{\frac{0.231*0.769}{432}}

The 95% confidence interval is 0.231 \pm 1.96\sqrt{\frac{0.231*0.769}{432}}

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