Question

Clarence wants to estimate the percentage of students who live more than three miles from the school. He wants to create a 98% confidence interval which has an error bound of at most 4%. How many students should be polled to create the confidence interval?

Answer 1

Answer:

Clarence should poll 847 to create the confidence interval.

Step-by-step explanation:

We have the following information:

1-α = 0.98

e = 0.04

p = unknown

The value of the sample (n) can be calculated using the following formula:

$n=\frac{z^{2}*p*(1-p) }{e^{2} }$

First step: obtain the value of p

Since we're not given a value of p, we'll have to use the value that maximizes the sample. The value that maximizes the sample is always p = 0.5.

Second step: obtain the value of z

Since the confidence interval is going to be of 98%, you’ll have 1% of the distribution in each tail. Therefore the z values we’re looking for are the cumulative values of the z distribution up to 1% or up to 99%. Either value will work, but you’ll just need to use one of them.

Looking up on any z-table, the value of a 99% area in a z distribution is of z = 2.327.

Third step: replace in the formula

$n=\frac{z^{2}*p*(1-p) }{e^{2} } = \frac{2.327^{2}*0.5*(1-0.5) }{0.04^{2}}=846.08$

Since we're looking for the number of people that should be polled, we cannot have decimals in the answer. Therefore we round up to 847 students.

Answer 2

Using the z-distribution, as we are working with a proportion, it is found that 847 students should be polled to create the confidence interval.

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.

The margin of error is given by:

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

98% confidence level, hence$\alpha = 0.98$, z is the value of Z that has a p-value of $\frac{1+0.98}{2} = 0.99$, so $z = 2.327$.

In this problem:

• There is no estimate of the proportion, hence $\pi = 0.5$.

• The margin of error is of M = 0.04.

To find the sample size, we solve for n, hence:

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

$0.04 = 2.327\sqrt{\frac{0.5(0.5)}{n}}$

$0.04\sqrt{n} = 2.327 \times 0.5$

$\sqrt{n} = \frac{2.327 \times 0.5}{0.04}$

$(\sqrt{n})^2 = \left(\frac{2.327 \times 0.5}{0.04}\right)^2$

$n = 846.1$

Rounding up, 847 students should be polled to create the confidence interval.

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