Practice 2 Solutions

Use the following scenario for problems 1–3.

A farming co-op is testing a new fertilizer to estimate the average yield per acre of corn. They randomly select and measure the yield from 81 acres. Based on decades of records for this region, the population standard deviation for corn yield is known to be 8.5 bushels/acre.

Estimate the co-op’s yield for a 95% confidence level.

$\begin{matrix} z = 1.960 & E = z \cdot \frac{σ}{\sqrt{n}} \\ σ = 8.5 & E = 1.96 \cdot (\frac{8.5}{\sqrt{81}}) \\ n = 81 & E = 1.8511 \end{matrix}$

Sample: The co-op can be 95% confident that their estimate will be no more than 1.85 bushels/acre from the population mean.

Determine the confidence interval when the sample mean yield is 150 bushels/acre.

$\begin{array}{rcl} \bar{x} - E & \leq & μ \leq \bar{x} + E \\ 150 - 1.85 & \leq & μ \leq 150 + 1.85 \\ 148.15 & \leq & μ \leq 151.85 \end{array}$

Sample: The co-op can be 95% confident that the population mean will be between 148.15 and 151.85 bushels/acre.

The co-op surveyed 162 community members to determine if the farming co-op had a positive, neutral, or negative perception. The results showed that 84% of the community had a positive perception of the farming co-op. Using a margin of error of $\pm 7.86 %$ , find the interval.

$\begin{array}{l} 84 - 7.86 = 76.14 % \\ 84 + 7.86 = 91.86 % \end{array}$

Sample: It is likely that between 76.14% and 91.86% of the community members have a positive view of the farming co-op.

Use the following scenario for problems 4–7. 

An environmental group is studying the average pH level of a large lake. Using historical data, they know the standard deviation is a pH of 0.5 parts per unit and are using a 95% confidence level. 

If the group takes an initial random sample of 50 water measurements, calculate the maximum error of the estimate.

$\begin{matrix} z = 1.960 & E = z \cdot \frac{σ}{\sqrt{n}} \\ σ = 0.5 & E = 1.96 \cdot (\frac{0.5}{\sqrt{50}}) \\ n = 50 & E = 0.1386 \end{matrix}$

Sample: The environmental group can be 95% confident that their estimate will be no more than 0.14 parts per unit from the population mean.

A new member of the environmental group suggested taking 200 samples from the lake. Without making calculations, why would a new member make this suggestion?

Sample: As the sample size increases, the maximum error of the estimate will decrease.

How many times larger is the new sample size?

$\frac{200}{50} = 4$

The new sample size is four times larger.

The value of E is 0.0692 for 200 lake samples. How many times smaller is the new value of E?

$\frac{0.1386}{0.0693} = 2$

The new value of E is two times smaller.

Note

You must quadruple the sample size to be twice as precise.

Use the following scenario for problems 8–10. 

A+ Service Company decides to estimate the average hold time for customers calling their support line. They want to be 95% confident that their estimate is within 0.5 minutes (30 seconds) of the true average. From prior data, the population standard deviation for hold time is 1.8 minutes. 

What is the minimum sample size of customer calls A+ Service Company needs to measure to meet this requirement?

$\begin{matrix} z = 1.960 & E = z \cdot \frac{σ}{\sqrt{n}} \\ σ = 1.8 & 0.5 = 1.96 \cdot (\frac{1.8}{\sqrt{n}}) \\ E = 0.5 & \frac{0.5}{1.96} = \frac{1.8}{\sqrt{n}} \\ 0.5 \sqrt{n} = 1.96 \cdot 1.8 \\ {(\sqrt{n})}^{2} = {(\frac{1.96 \cdot 1.8}{0.5})}^{2} \\ n = 49.787 \\ n = 50 \end{matrix}$

Sample: To be 95% confident in hold time, A+ Service Company needs to survey a minimum of 50 customers.

After the customers were surveyed, the sample mean for hold time was calculated to be 2.3 minutes. Determine the confidence interval.

$\begin{array}{rcl} \bar{x} - E & \leq & μ \leq \bar{x} + E \\ 2.3 - 0.5 & \leq & μ \leq 2.3 + 0.5 \\ 1.8 & \leq & μ \leq 2.8 \end{array}$

Sample: A+ Service Company can be 95% confident that the population mean will be between 1.8 and 2.8 minutes.

At the end of every customer interaction, clients are asked to rate their experience with A+ Service Company on a scale of 1 (terrible) to 5 (wonderful). A random sample of 340 responses was collected, revealing that 76.5% of customers rated their experience as 4 or higher. Using a margin of error of $\pm 5.42 %$ , calculate the interval.

$\begin{array}{l} 76.5 - 5.42 = 71.08 % \\ 76.5 + 5.42 = 81.92 % \end{array}$

Sample: It is likely that between 71.08% and 81.92% of customers will rate A+ Service Company at 4 or higher.

Use the following scenario for problems 11–12.

Civil engineers are measuring the length of a bridge span and must have a high degree of certainty in their measurements. They take 196 independent measurements using instruments with a known population standard deviation of 0.8 inches. They must be 99% confident in their estimation of the true length.

Determine the maximum error of the estimate if the engineers must be 99% confident in their estimate of the true length.

$\begin{matrix} z = 2.576 & E = z \cdot \frac{σ}{\sqrt{n}} \\ σ = 0.8 & E = 2.576 \cdot (\frac{0.8}{\sqrt{196}}) \\ n = 196 & E = 0.1472 \end{matrix}$

Sample: The engineers can be 99% confident that the maximum error of the estimate is 0.15 inches.

Note

For bridge building, precision is critical. An error of less than a quarter of an inch, even at the high 99% confidence level, shows that the large sample size $(n = 196)$ is doing a lot of heavy lifting to keep the estimate tight and reliable.

The engineers found the sample mean bridge length to be 89.9 feet (1,078.8 inches) from the sample measurements. Determine the population mean with 99% confidence.

$\begin{array}{rcl} \bar{x} - E & \leq & μ \leq \bar{x} + E \\ 1078.8 - 0.1472 & \leq & μ \leq 1078.8 + 0.1472 \\ 1078.65 & \leq & μ \leq 1078.95 \end{array}$

Sample: The engineers can be 99% confident that the population mean will be between 1,078.65 and 1,078.95 inches.

Use the following scenario for problems 13–14.

A food safety inspector is estimating the average sodium content (in mg) of a new frozen meal. They need to be very precise, requiring an estimate that is within 5 mg of the true mean. The known population standard deviation for sodium content in this type of meal is 18 mg.

What is the minimum sample size of meals they need to test to guarantee a 99% confidence level?

$\begin{matrix} z = 2.576 & E = z \cdot \frac{σ}{\sqrt{n}} \\ σ = 18 & 5 = 2.576 \cdot (\frac{18}{\sqrt{n}}) \\ E = 5 & \frac{5}{2.576} = \frac{18}{\sqrt{n}} \\ 5 \sqrt{n} = 2.576 \cdot 18 \\ {(\sqrt{n})}^{2} = {(\frac{2.576 \cdot 18}{5})}^{2} \\ n = 85.9997 \\ n = 86 \end{matrix}$

86 meals

What should the food inspector do if the value of E needs to be more precise (smaller)?

Sample: The sample size can be increased to reduce the value of E.

Sample: The confidence level can be changed to 90% or 95% because these are more precise (closer to the true population mean).