Question

A business student is interested in estimating the 95% confidence interval for the proportion of students who bring laptops to campus.
He wishes a precise estimate and is willing to draw a large sample that will keep the sample proportion within five percentage points
of the population proportion. What is the minimum sample size required by this student, given that no prior estimate of the population
proportion is available?
(You may find it useful to reference the z table. Round intermediate calculations to at least 4 decimal places
and "2" value to 3 decimal places. Round up your answer to the nearest whole number.)

Answer 1

Using the z-distribution, the minimum sample size required by this student is of 385.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

The margin of error is:

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

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.

We want a margin of error of M = 0.05, with no prior estimate, hence $\pi = 0.5$, then we have to solve for n to find the minimum sample size.

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

$0.05 = 1.96\sqrt{\frac{0.5(0.5)}{n}}$

$0.05\sqrt{n} = 1.96(0.5)$

$\sqrt{n} = 1.96 \times 10$

$(\sqrt{n})^2 = (1.96 \times 10)^2$

n = 384.16

Rounding up, the minimum sample size is of 385.

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

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