Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 195, x = 126; 95% confidence

Answer 1

Using the z-distribution, the 95% confidence interval for the population proportion is (0.5791, 0.7133).

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

In this problem, we have a 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so the critical value is z = 1.96.

The other parameters are given as follows:

$n = 195, \overline{p} = \frac{126}{195} = 0.6462$

Hence the bounds of the interval are:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6462 - 1.96\sqrt{\frac{0.6462(0.3538)}{195}} = 0.5791$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6462 + 1.96\sqrt{\frac{0.6462(0.3538)}{195}} = 0.7133$

More can be learned about the z-distribution at brainly.com/question/25890103

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