Practice 1 Solutions

1. Calculate the maximum error of the estimate if they aim for a 90% confidence level in their study. Explain.

$\begin{array}{ccc}z=1.645& & E=z·\frac{s}{\sqrt{n}}\\ \sigma =6.2& & E=1.645\left(\frac{6.2}{\sqrt{250}}\right) \\ n=250& & E=0.6450\end{array}$

Sample: The Department of Transportation can be 90% confident that its estimate will be no more than 0.65 minutes from the population mean.

2. Calculate the maximum error of the estimate if the random sample increased to 500 commuters. Explain.

$\begin{array}{ccc}z=1.645& & E=z·\frac{s}{\sqrt{n}}\\ \sigma =6.2& & E=1.645\left(\frac{6.2}{\sqrt{500}}\right) \\ n=500& & E=0.4561\end{array}$

Sample: For 500 commuters, the Department of Transportation can be 90% confident that its estimate will be no more than 0.46 minutes from the population mean.

3. Why is the error reduced when the sample increases?

Sample: As the sample size increases, the value of E decreases because larger samples are closer to the true population size.

4. If the sample mean for 500 commuters is 38 minutes, determine the confidence interval (CI).

$\begin{array}{rcl}\overline{x}–E & \le & \mu \le \overline{x}+E\\ 38–0.4561& \le & \mu \le 38+0.4561\\ 37.5439& \le & \mu \le 38.4561\end{array}$

Sample: You can be 90% certain that the population mean will be between 37.54 and 38.46 minutes.

5. Calculate the maximum error of the estimate.

$\begin{array}{ccc}z=2.576& & E=z·\frac{s}{\sqrt{n}}\\ \sigma =15& & E=2.576\left(\frac{15}{\sqrt{64}}\right) \\ n=64& & E=4.83\end{array}$

$E=4.83$

6. Why would the school use a 99% confidence level instead of a 90% confidence level?

Sample: A 99% confidence level gives the school a larger target, so it is more likely to include the population mean in the sample.

7. Lakeforest School District surveyed 136 third-grade parents about the reading program their children use at school. Approximately 11% of parents were dissatisfied with the program. Using a $\pm 8.4\%$ margin of error, find the interval that most likely contains the population parameter for satisfied parents.

$$\begin{gathered} \begin{array}{l}100–11=89\% \\ 89–8.4=80.6\%\\ 89+8.4=97.4\%\end{array} \end{gathered}$$

Sample: It is likely that between 80.6% and 97.4% of parents are satisfied with the reading program.

8. What is the minimum number of parts to be tested to match the confidence level?

$$\begin{gathered} \begin{array}{ccc}z=1.960& & E=z·\frac{\sigma }{\sqrt{n}}\\ \sigma =0.05& & 0.01=1.960·\left(\frac{0.05}{\sqrt{n}}\right) \\ E=0.01& & \frac{0.01}{1.96}=\frac{0.05}{\sqrt{n}}\\ & & 0.01\sqrt{n}=1.96·0.05\\ & & {\left(\sqrt{n}\right)}^2={\left(\frac{1.96·0.05}{0.01}\right)}^2\\ & & n=96.04 \\ n=97\end{array} \end{gathered}$$

Sample: The automobile parts supplier must sample a minimum of 97 parts to meet a 0.01 mm requirement.

9. Determine the population mean from the random sample of pistons when the sample mean is 95.5 mm.

$\begin{array}{rcl}\overline{x}–E & \le & \mu \le \overline{x}+E\\ 95.5–0.01& \le & \mu \le 95.5+0.01\\ 95.49& \le & \mu \le 95.51\end{array}$

Sample: The automobile parts supplier can be 95% confident that the population mean will be between 95.49 mm and 95.51 mm.

10. Determine the maximum error of the estimate in days and hours.

$\begin{array}{ccccc}z=1.645& & E=z·\frac{\sigma }{\sqrt{n}}& & \\ \sigma =2.4& & E=1.645·\left(\frac{2.4}{\sqrt{120}}\right) & & 0.3604 \mathrm{day}\left(\frac{24 \mathrm{hr}}{\mathrm{day}}\right)\\ n=120& & E=0.3604& & 8.6 \mathrm{hr}\end{array}$

Sample: The maximum error of the estimate is 0.36 days (about 8.6 hours).

11. If the average stay of a randomly selected group of patients is 4.5 days, what is the population mean?

$\begin{array}{rcl}\overline{x}–E & \le & \mu \le \overline{x}+E\\ 4.5–0.3604& \le & \mu \le 4.5+0.3604\\ 4.1396& \le & \mu \le 4.8604\end{array}$

Sample: At a 90% confidence level, the population mean for general surgery patients will be between 4.14 and 4.86 days.

12. The hospital administrator decided to adjust the confidence level. Calculate the maximum error of the estimate for a 95% confidence level.

$\begin{array}{ccccc}z=1.960& & E=z·\frac{\sigma }{\sqrt{n}}& & \\ \sigma =2.4& & E=1.96·\left(\frac{2.4}{\sqrt{120}}\right) & & 0.4294 \mathrm{day}\left(\frac{24 \mathrm{hr}}{\mathrm{day}}\right)\\ n=120& & E=0.4294& & 10.3 \mathrm{hr}\end{array}$

Sample: The maximum error of the estimate is 0.43 days (about 10.3 hours).

13. If hospital employees are scheduled in 8-hour shifts, explain which confidence level should be used.

90% confidence level is 8.6 hours
95% confidence level is 10.3 hours

Sample: The 90% confidence level is more precise and closer to the shift length employees work. This would better align with scheduling decisions.

14. The same sample of patients was also given a discharge survey asking them to rate their hospital stay as positive, neutral, or negative. Of the 120 patients surveyed, 17% rated their stay as neutral. Find the interval when the margin of error is $\pm 9.13\%.$

$\begin{array}{l}17–9.13=7.87\%\\ 17+9.13=26.13\%\end{array}$

$7.87\%\le \mu \le 26.13\%$