Question

In a sample of 230 teenagers, 202 would like to have smaller class sizes at their school. Find the sample

Answer 1

Answer:

The sample proportion is 0.88.

The margin of error is of 0.04.

The interval likely to contain the true population proportion is (0.84,0.92).

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 $1 - \frac{\alpha}{2}$.

Margin of error:

The margin of error is of:

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

Sample of 230 teenagers, 202 would like to have smaller class sizes at their school.

This means that $n = 230$, and that the sample proportion is:

$\pi = \frac{202}{230} = 0.88$

The sample proportion is 0.88.

Standard 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 $Z = 1.96$.

Margin of error:

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

$M = 1.96\sqrt{\frac{0.88*0.12}{230}}$

$M = 0.04$

The margin of error is of 0.04.

Confidence interval:

Sample proportion plus/minus margin of error. So

0.88 - 0.04 = 0.84

0.88 + 0.04 = 0.92

The interval likely to contain the true population proportion is (0.84,0.92).