Question

order to estimate the average time spent on the computer terminals per student at a university, data were collected for a sample of 81 business students over a one week period. Assume the population standard deviation is 1.8 hours. With a 0.95 probability, the margin of error is approximately a. 0.20 b. 0.39 c. 1.96 d. 1.64

Answer 1

Using the z-distribution, as we have the standard deviation for the population, the margin of error is given by:

b. 0.39.

What is a t-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 sample.

The margin of error is given by:

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

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.

As for the other parameters, we have that $\sigma = 1.8, n = 81$.

Hence, the margin of error is given by:

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

$M = 1.96\frac{1.8}{\sqrt{81}}$

$M = 0.39$

Hence option b is correct.

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