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:

$\pi = 0.21$

The margin of error is:

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

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 $Z = 1.96$ .

What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.09?

A sample size of n is needed.

n is found when M = 0.09. So

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

$0.09 = 1.96\sqrt{\frac{0.21*0.79}{n}}$

$0.09\sqrt{n} = 1.96\sqrt{0.21*0.79}$

$\sqrt{n} = \frac{1.96\sqrt{0.21*0.79}}{0.09}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.21*0.79}}{0.09})^{2}$

$n = 78.68$

Rounding up to the nearest whole number.

A sample size of 79 is needed.

4 0

3 years ago

After how many audio books will both book clubs cost the same amount.

bulgar [2K]

Answer:

2 books

Step-by-step explanation:

10 + 5x = 16 + 2x

- 2x - 2x

10 + 3x = 16

-10 -10

3x = 6

/3 /3

4 0

2 years ago

Hey can u help me how to do this plz.

Svetach [21]

Okay so x represents the amount of tickets sold. So first, you’d add the $40 + $70 which equals $110. So x = $110.

Now we have to solve for y. Same thing but with different numbers. $200 + $260 = $460.
The answer is (110,460)
x. y.

4 0

2 years ago