Question

The Student Monitor surveys 1200 undergraduates from 100 colleges semiannually to understand trends among college students. Recently, the Student Monitor reported that the average amount of time spent per week on the Internet was 19.0 hours. You suspect that this amount is far too small for your campus and plan a survey. You feel that a reasonable estimate of the standard deviation is 10.0 hours. What sample size is needed so that the expected margin of error of your estimate is not larger than one hour for 95% confidence

Answer 1

Answer:

A sample size of 385 is needed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$.

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

You feel that a reasonable estimate of the standard deviation is 10.0 hours.

This means that $\sigma = 10$

What sample size is needed so that the expected margin of error of your estimate is not larger than one hour for 95% confidence?

A sample size of n is needed. n is found when M = 1. So

$M = z\frac{\sigma}{\sqrt{n}}$

$1 = 1.96\frac{10}{\sqrt{n}}$

$\sqrt{n} = 1.96*10$

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

$n = 384.16$

Rounding up:

A sample size of 385 is needed.