Chapter 23, Problem 14E

(a)

To determine

To explain what assumptions must you make in order to use these statistics for inference.

Explanation of Solution

It is given in the question that hoping to lure more shoppers downtown a city builds a new parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period, daily fees collected averaged $\$126$ , with a standard deviation of $\$15$ . Thus, the conditions for creating a t -interval to be checked are:

Random: It is satisfied because assuming that the daily fees were randomly selected.

Independent condition: It is satisfied because it can be assumed that the $44$ daily fees are less than $10\%$ of the population, thus the population should contain more than $440$ daily fees..

Normal condition: It is satisfied because the sample size of $\text{44 is 30}$ or more and thus the sampling distribution of the sample mean is approximately normal.

Thus, we have that all the three conditions are met.

(b)

To determine

To find a $90\%$ confidence interval for mean daily income this parking garage will generate.

Answer to Problem 14E

The confidence interval is $(122.1919,129.8081)$ .

Explanation of Solution

It is given in the question that hoping to lure more shoppers downtown a city builds a new parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period, daily fees collected averaged $\$126$ , with a standard deviation of $\$15$ . And all conditions are met.

It is given:

  $\begin{array}{l}n=44\\ \overline{x}=126\\ s=15\end{array}$

Now we will find the t -value by looking in the row starting with degrees of freedom. The degree of freedom is as:

  $\begin{array}{l}df=n-1\\ =44-1\\ =43\end{array}$

Thus, the t -value with $90\%$ confidence interval be:

  $t_{\alpha /2}=1.684$

The confidence interval is calculated as:

  $\begin{array}{l}\overline{x}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =126-1.684\times \frac{15}{\sqrt{44}}\\ =122.1919\\ \overline{x}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =126+1.684\times \frac{15}{\sqrt{44}}\\ =129.8081\end{array}$

Thus the confidence interval is $(122.1919,129.8081)$ .

(c)

To determine

To interpret the confidence interval in this context.

Explanation of Solution

It is given in the question that hoping to lure more shoppers downtown a city builds a new parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period, daily fees collected averaged $\$126$ , with a standard deviation of $\$15$ . And all conditions are met.

It is given:

  $\begin{array}{l}n=44\\ \overline{x}=126\\ s=15\end{array}$

And the confidence interval is $(122.1919,129.8081)$ . Thus, it means that we are $90\%$ confident that the true average daily fee is between $\text{\$}122.1919\text{ and \$}129.8081$ .

(d)

To determine

To explain what “ $90\%$ confidence” means in this context.

Explanation of Solution

It is given in the question that hoping to lure more shoppers downtown a city builds a new parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period, daily fees collected averaged $\$126$ , with a standard deviation of $\$15$ . And all conditions are met.

It is given:

  $\begin{array}{l}n=44\\ \overline{x}=126\\ s=15\end{array}$

And the confidence interval is $(122.1919,129.8081)$ . The term “ $90\%$ confidence” means in this context that the confidence interval will contain the true population mean in $90\%$ of all possible samples.

(e)

To determine

To explain do you think the consultant was correct, based on your confidence interval.

Answer to Problem 14E

No, the consultant was not correct.

Explanation of Solution

It is given in the question that hoping to lure more shoppers downtown a city builds a new parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period, daily fees collected averaged $\$126$ , with a standard deviation of $\$15$ . And all conditions are met.

It is given:

  $\begin{array}{l}n=44\\ \overline{x}=126\\ s=15\end{array}$

And the confidence interval is $(122.1919,129.8081)$ . Thus, the confidence interval does not contain $\$130$ and thus this data suggest that the average is not $\$130$ , which means that the consultant appears to be not correct.