Test 25 (Lessons 49–50): Confidence, Error, and Experiments Solutions

For problems 1–3, use the following scenario.

Bean & Brew Cafe wants to estimate the average time customers spend at the counter. They want to be 95% confident that the estimate is within 0.3 minutes during the morning rush. The computer system reports the standard deviation is 1.3 minutes.

What is the minimum number of customer encounters needed to obtain the confidence level?

$\begin{matrix} z = 1.96 & E = z \cdot \frac{s}{\sqrt{n}} \\ E = 0.3 & 0.3 = 1.96 \cdot \frac{1.3}{\sqrt{n}} \\ s = 1.3 & \frac{0.3 \sqrt{n}}{0.3} = \frac{2.548}{0.3} \\ {(\sqrt{n})}^{2} = {(8.49 \bar{3})}^{2} \\ n = 72.14 \end{matrix}$

Bean & Brew Cafe must have a minimum of 73 customer encounters to obtain a 95% confidence level.

After meeting the minimum requirement, the sample mean time was 3.7 minutes. Determine the confidence interval at the 95% level.

$\begin{array}{rcl} \bar{x} - E & \leq & μ \leq \bar{x} + E \\ 3.7 - 0.3 & \leq & μ \leq 3.7 + 0.3 \\ 3.4 & \leq & μ \leq 4.0 \end{array}$

Bean & Brew Cafe asked its customers how satisfied they were with the Cafe after their order. Customers reported they were 92% satisfied with a margin of error ±1.3%.

Determine the interval that most likely contains the population average.

$92 % - 1.3 % = 90.7 % 92 % + 1.3 % = 93.3 % 90.7 % \leq μ \leq 93.3 %$

For problems 4–6, use the following scenario.

A science class conducted a blind taste test with 140 participants to determine if people can taste the difference between bottled water and the school’s water fountain water. A randomized simulation with 10,000 trials resulted in an observed difference of 0.4 points. The standard deviation of the simulated difference was 1.38 points.

Group n = 70	Mean Taste Rating (1–10)
Treatment: bottled water	6.2
Control: water fountain water	5.8

Additional Considerations:

The water fountain water is filtered.
The science class made sure all water was the same temperature before the taste test.
A typical student drinks the equivalent of 4 bottles of water daily.
A student can purchase a bottle of water for $1.50 or a reusable water bottle for $12.

Calculate the z-score for the observed difference. Explain the statistical significance at the 5% level.

$6.2 - 5.9 = 0.3 z = \frac{0.3}{1.38} = 0.217$

Sample: Because the z-score is well below 1.96, there is no statistical significance.

Do the results prove that people can correctly identify the bottled water?

Sample: No, the results show that people scored the taste of bottled water 0.4 higher on average. It does not show they can identify the type of water.

Calculate how much money a student would spend on bottled water for a 180-day school year. Compare this to the cost of a reusable water bottle.

$4 ($ 1.50) = $ 6.00 / day ($ 6) (180) = $ 1, 080 / school year $ 1, 080 - $ 12 = $ 1, 068 savings$

Based on the statistical evidence and cost comparison, what type of water would you recommend students drink?

Sample: I would recommend that students drink the water from the fountain because they would save over $1,000 per year.

For problems 8–10, use the scenario below.

Fit-4-Life Gym wants to estimate the number of members expected to renew their annual gym membership. The gym randomly surveys 225 members whose membership is expiring soon. From previous years, Fit-4-Life has calculated the renewal standard deviation to be 29.2 members.

Calculate and explain the maximum error of the estimate for the 90% level, $z = 1.645 .$

$\begin{matrix} z = 1.645 & E = z \cdot \frac{s}{\sqrt{n}} \\ s = 29.2 & E = 1.645 \cdot \frac{29.2}{\sqrt{225}} \\ n = 225 & E = 3.2022 \end{matrix}$

Fit-4-Life Gym can be 90% confident that its estimate will be no more than 3.2022 (about 4) people away from the population mean.

Determine a 99% confidence interval if the average renewal rate is 82% and $E = 5.01 .$ Explain.

$\begin{array}{rcl} \bar{x} - E & \leq & μ \leq \bar{x} + E \\ 82 - 5.01 & \leq & μ \leq 82 + 5.01 \\ 76.99 & \leq & μ \leq 87.01 \end{array}$

You can be 99% certain that the average renewal rate for gym members will be between 76.99% and 87.01%.

If the gym manager wants to be as accurate as possible, which is a better estimate? Explain.

The 90% level is a better estimate because it is a more precise representation of confidence.