Mean Confidence Worksheet
Mean Confidence Lab
Confidence Intervals for Population Means
Researcher
Date
Lab Procedures
Complete the following five statistical problems. For all calculations involving standard normal distributions, use \( z \)-values rounded to three decimal places where applicable (e.g., \( 1.960 \) for 95% confidence). Show your work clearly for each calculation.
1
A researcher is designing a study and requires a high level of precision. Find the critical value \( z^* \) for a 98% confidence level. Round your answer to three decimal places.
2
A marine biologist measures the weight of 49 sea turtles. The population standard deviation is known to be \( \sigma = 4.2 \) kg. Calculate the margin of error for a 95% confidence interval for the true mean weight of the population.
3
A manufacturing plant wants to estimate the mean weight of produced bolts. They want to be 99% confident that the sample mean is within 0.15 grams of the true population mean. If the population standard deviation is \( \sigma = 0.8 \) grams, what is the minimum sample size (\( n \)) required?
4
A random sample of 64 lightbulbs has a mean lifespan of 1,200 hours. Assuming the population standard deviation is \( \sigma = 80 \) hours, calculate the lower bound of a 90% confidence interval for the population mean lifespan.
5
If a population has a standard deviation of 18, what is the minimum number of samples that need to be averaged in order to be 95% confident that the average of the means is within 4 of the true mean?
A 9
B 36
C 78
D 11
Mean Confidence Step Key
Step-by-Step Solutions
Mean Confidence Lab Instruction Manual
Teacher Resource
Key Formulas
Confidence Interval
\[ \bar{x} \pm z^* \left( \frac{\sigma}{\sqrt{n}} \right) \]
Sample Size Calculation
\[ n = \left( \frac{z^* \cdot \sigma}{E} \right)^2 \]
Problem 1: 98% Critical Value
Step 1: Identify the area in the tails. For 98% confidence, \( \alpha = 1 - 0.98 = 0.02 \).
Step 2: Split the area into two tails: \( \alpha/2 = 0.01 \).
Step 3: Find the z-score that corresponds to a cumulative area of \( 1 - 0.01 = 0.99 \).
Answer: \( z^* = 2.326 \)
Problem 2: Marine Biologist MOE
Given: \( n = 49 \), \( \sigma = 4.2 \), \( CL = 95\% \Rightarrow z^* = 1.96 \).
Formula: \( E = z^* \left( \frac{\sigma}{\sqrt{n}} \right) \)
Calculation: \( E = 1.96 \left( \frac{4.2}{\sqrt{49}} \right) = 1.96 \left( \frac{4.2}{7} \right) = 1.96 (0.6) = 1.176 \).
Answer: Margin of Error = 1.176 kg
Problem 3: Bolt Manufacturing Sample Size
Given: \( CL = 99\% \Rightarrow z^* = 2.576 \), \( E = 0.15 \), \( \sigma = 0.8 \).
Formula: \( n = \left( \frac{2.576 \cdot 0.8}{0.15} \right)^2 \)
Calculation: \( n = \left( \frac{2.0608}{0.15} \right)^2 = (13.73866...)^2 = 188.75 \).
Note: Always round up to the next whole integer for sample size.
Answer: n = 189 bolts
Problem 4: Lightbulb Lower Bound
Given: \( \bar{x} = 1200 \), \( n = 64 \), \( \sigma = 80 \), \( CL = 90\% \Rightarrow z^* = 1.645 \).
MOE: \( E = 1.645 \left( \frac{80}{\sqrt{64}} \right) = 1.645 (10) = 16.45 \).
Lower Bound: \( \bar{x} - E = 1200 - 16.45 = 1183.55 \).
Answer: 1,183.55 hours
Problem 5: Multiple Choice Sample Size
Given: \( \sigma = 18 \), \( CL = 95\% \Rightarrow z^* = 1.96 \), \( E = 4 \).
Formula: \( n = \left( \frac{1.96 \cdot 18}{4} \right)^2 = \left( \frac{35.28}{4} \right)^2 = (8.82)^2 = 77.79 \).
Result: Rounding up gives \( n = 78 \).
Answer: Choice C (78)
