91影视

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

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

a.

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

$CI=\stackrel{炉}{x}-E<渭<\stackrel{炉}{x}+E$

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

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

Therefore, the confidence interval can be expressed as follows.

$$\begin{gathered} \stackrel{炉}{x}-E<渭<\stackrel{炉}{x}+E \\ 13.05<渭<22.15 \end{gathered}$$

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

b.

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

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

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

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

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

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

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

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.