Question

300 high school students were asked how many hours of TV they watch
per day. The mean was 2 hours, with a standard deviation of 0.5. Using a
90% confidence level, calculate the maximum error of estimate.

Answer 1

Using the z-distribution, it is found that the maximum error of the estimate is of 0.0475 hours.

What is a z-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

The margin of error is given by:

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

For this problem, the values of the parameters are given as follows:

$\sigma = 0.5, n = 300$

We have a 90% confidence level, hence$\alpha = 0.9$, z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$, so the critical value is z = 1.645.

Thus the maximum error of the estimate in hours is found as follows:

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

$M = 1.645\frac{0.5}{\sqrt{300}}$

M = 0.0475 hours

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

#SPJ1