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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 865, \pi = \frac{408}{865} = 0.4717$

95% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4717 - 1.96\sqrt{\frac{0.4717*0.5283}{865}} = 0.4384$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4717 + 1.96\sqrt{\frac{0.4717*0.5283}{865}} = 0.5050$

The 95% confidence interval for the true proportion of all voters in the state who favor approval is (0.4384, 0.5050).

6 0

3 years ago

Solve the equation or formula for the indicated variable.

alex41 [277]

Subtract 3 both sides first. R-3=ts²

next, divide both sides by t, (R-3)/t=s²

finally, take the square root of both sides

√(R-3)/t = s. 2nd choice.

3 0

3 years ago

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