Interval Investigator Worksheet
Interval Investigator
Confidence Intervals for Means & Margin of Error
Subject: Statistics
Date:
Researcher:
Mission Objective
Master the art of interval estimation. You will calculate margins of error, predict how shifting variables impact your results, and construct confidence intervals for population means from both summary statistics and raw data sets.
1
Margin of Error Mechanics
Calculate the margin of error \(E\) using the formula: \(E = z^* \cdot \frac{\sigma}{\sqrt{n}}\). Round to two decimal places.
1.1
A population has a standard deviation of \(\sigma = 12.5\). A researcher draws a sample of size \(n = 100\). Find the margin of error for a 95% confidence interval for the mean (\(z^* = 1.96\)).
1.2
Calculate the margin of error for a 99% confidence interval (\(z^* = 2.576\)) given a sample size of \(n = 64\) and a population standard deviation of \(\sigma = 24\).
1.3
A biologist measures the wing length of 40 birds. The population standard deviation is known to be \(\sigma = 3.2\) cm. Determine the margin of error for a 90% confidence interval (\(z^* = 1.645\)).
2
The Logic of Error
Analyze the conceptual relationships within the interval formula.
2.1In a study regarding coffee consumption, the current margin of error is 2.5 units. If the researcher increases the sample size while keeping the confidence level and standard deviation constant, what happens to the margin of error?
A. It increases
B. It decreases
C. No change
D. Not enough info
2.2A marketing firm wants to be more certain about their results, so they change their confidence level from 90% to 99%. Assuming the sample data stays the same, how will this change affect the width of their confidence interval?
A. The interval widens
B. The interval narrows
C. The width remains the same
D. The center shifts
2.3Which of the following experimental designs would produce the most precise (narrowest) margin of error, assuming \(\sigma\) is constant?
A. 99% Confidence Level with \(n = 50\)
B. 95% Confidence Level with \(n = 50\)
C. 95% Confidence Level with \(n = 200\)
D. 90% Confidence Level with \(n = 200\)
3
Constructing the Interval
Calculate the confidence interval using \(\bar{x} \pm E\). Round results to one decimal place.
3.1
A sample of 45 commute times shows a mean of \(\bar{x} = 32.4\) minutes. If the population standard deviation is \(\sigma = 6.2\) minutes, construct a 95% confidence interval for the average commute time of all workers.
Show work: Find \(E\), then construct interval
3.2
The prices for 5 random laptops at a store are: $899, $1250, $945, $1100, $1050. Given a population standard deviation of \(\sigma = 150\), construct a 90% confidence interval (\(z^* = 1.645\)) for the mean price.
Calculate \(\bar{x}\) first
3.3
A nutritionist finds that 38 light-yogurt cups have an average of \(\bar{x} = 142\) calories. Assuming \(\sigma = 18\) calories, find the 98% confidence interval (\(z^* = 2.33\)) for the mean calories of all cups.
4
Planning the Study
Determine the minimum sample size \(n\) required. Always round up to the next whole number.
4.1
A scientist wants to estimate the mean weight of adult squirrels within 0.5 ounces with 95% confidence. If \(\sigma = 3.8\) ounces, what is the minimum number of squirrels that must be weighed?
4.2
A city planner needs to estimate the mean water usage per household within 2 gallons. With \(\sigma = 15\) gallons and a 99% confidence level (\(z^* = 2.576\)), find the required sample size.
4.3
If the population standard deviation for test scores is 14, how large a sample is needed to be 90% confident (\(z^* = 1.645\)) that the sample mean is within 3 points of the true population mean?
5
Interpretation Mastery
Apply your findings to make statistical claims.
5.1
Three confidence intervals were constructed from the same sample of data:
- Interval A: \(42.5 < \mu < 47.5\)
- Interval B: \(41.2 < \mu < 48.8\)
- Interval C: \(43.1 < \mu < 46.9\)
If the confidence levels used were 90%, 95%, and 99%, identify which interval corresponds to the 99% confidence level and justify your reasoning.
5.2
A researcher constructs a 95% confidence interval for the mean height of a plant species: \(12.4 \text{ cm} < \mu < 15.6 \text{ cm}\). Is it reasonable to conclude that the population mean could be 16.0 cm? Explain why or why not.
5.3
A study found a 90% confidence interval for the mean number of hours students sleep to be \((6.8, 7.4)\). If the margin of error was 0.3, what was the sample mean \(\bar{x}\) used in this study?
Interval Investigator Answer Key
Answer Key
Interval Investigator: Confidence Intervals
Teacher Resource
Numerical Results
| # | Solution / Answer | # | Solution / Answer |
|---|
| 1.1 | 2.45 | 3.1 | 30.6 < μ < 34.2 |
| 1.2 | 7.73 | 3.2 | 938.5 < μ < 1159.1 |
| 1.3 | 0.83 | 3.3 | 135.2 < μ < 148.8 |
| 2.1 | B (Decreases) | 4.1 | 222 squirrels |
| 2.2 | A (Widens) | 4.2 | 374 households |
| 2.3 | D (90%, n=200) | 4.3 | 59 students |
Section 5 Explanations
5.1 Determining the 99% Interval
The correct answer is Interval B (\(41.2 < \mu < 48.8\)). Justification: For a fixed sample, as the confidence level increases, the critical value \(z^*\) increases, which increases the margin of error. Therefore, the 99% confidence level must produce the widest interval of the three options.
5.2 Statistical Reasoning
Conclusion: No, it is not reasonable. Justification: The 95% confidence interval \((12.4, 15.6)\) represents the range of values that are plausible for the population mean. Since 16.0 cm falls outside this interval, we do not have evidence to suggest it could be the true population mean at this level of confidence.
5.3 Finding the Point Estimate
The sample mean \(\bar{x}\) is 7.1 hours. Calculation: The sample mean is the exact midpoint of the confidence interval. \(\frac{6.8 + 7.4}{2} = 7.1\) or \(6.8 + 0.3 = 7.1\).
Grading Key & Common Misconceptions
Rounding Errors:
Ensure students round up for sample size (\(n\)) problems, even if the decimal is small (e.g., 58.1 becomes 59).
Formula Mixups:
Watch for students using \(\sigma\) instead of \(\frac{\sigma}{\sqrt{n}}\) when calculating the margin of error.
