Question

In Exercises 1–3, refer to the accompanying screen display that results from the Verizon airport data speeds (Mbps) from Data Set 32 “Airport Data Speeds” in Appendix B. The confidence level of 95% was used

Airport Data Speeds Refer to the accompanying screen display.

a. Express the confidence interval in the format that uses the “less than” symbol. Given that the original listed data use one decimal place, round the confidence interval limits accordingly.

b. Identify the best point estimate of and the margin of error.

c. In constructing the confidence interval estimate of , why is it not necessary to confirm that the sample data appear to be from a population with a normal distribution?

Short answer

a. The confidence interval is $13.05\text{Mbps}<\mu <22.15\text{Mbps}$.

b. The best point estimate of $\mu$is equal to 17.60 Mbps. The value of the margin of error is equal to 4.55 Mbps.

c. There is no need to confirm that the sample data appears to be from a population with a normal distribution because the sample size is greater than 30.

Step-by-step solution

Given information

The following values are given for the data on Verizon airport data speeds:

• Confidence interval: (13.046, 22.15)
• Sample mean $\left(\overline{x}\right)$: 17.598
• Sample size (n): 50

Expression of the confidence interval

a.

The following expression can be used to denote the confidence interval:

$CI=\overline{x}-E<\mu <\overline{x}+E$

Here, $\overline{x}-E$ is the lower limit, and $\overline{x}+E$ is the upper limit.

Thus, the value of 13.046 is equal to $\overline{x}-E$, and the value of 22.15 is equal to $\overline{x}+E$.

Therefore, the confidence interval can be expressed as follows.

$$\begin{gathered} \overline{x}-E<\mu <\overline{x}+E \\ 13.05<\mu <22.15 \end{gathered}$$

Thus, the confidence interval is $13.05\text{Mbps}<\mu <22.15\text{Mbps}$.

The best point estimate of  μ and the margin of error

b.

It is known that the best point estimate of the population mean is the sample mean.

Here, the value of $\overline{x}$ is equal to 17.60 Mbps.

Therefore, the best point estimate of $\mu$ is equal to 17.60 Mbps.

The value of the margin of error is computed as follows.

$$\begin{gathered} \overline{x}-E=13.05 \\ E=\overline{x}-13.05 \\ =17.60-13.05 \\ =4.55 \end{gathered}$$

Thus, the value of the margin of error is equal to 4.55 Mbps.

Requirement for constructing the confidence interval

c.
One of the requirements that need to be fulfilled in order to compute the confidence interval is as follows.

• The sample should be from a normally distributed population, or the sample size should be greater than 30.

Therefore, as the sample size equal to 50 is greater than 30, there is no need to confirmthat the sample data appear to be from a population with a normal distribution.