Module 5 Practice Problems
20 Comprehensive Problems • Covers All Module 5 Topics
How to Use These Practice Problems
- Work through problems on paper before checking solutions
- Use the hints if you're stuck
- Problems are organized by topic and difficulty
- Show all your work - understanding the process is key!
Problems 1-5: Parameters vs. Statistics & Sampling Distribution Basics
Problem 1 Identifying Parameters and Statistics Easy
For each scenario, identify the parameter and the statistic:
(a) A university surveys 300 of its 15,000 students. The average GPA of the 300 students is 3.2.
(b) The Department of Health wants to know what percentage of all California adults have health insurance. They survey 5,000 adults and find 88% have insurance.
Solution:
(a)
- Population: All 15,000 students
- Parameter: μ = true average GPA of all 15,000 students (unknown)
- Sample: 300 students surveyed
- Statistic: x̄ = 3.2 (average GPA of the 300 students)
(b)
- Population: All California adults
- Parameter: p = true proportion of all California adults with insurance (unknown)
- Sample: 5,000 adults surveyed
- Statistic: p̂ = 0.88 or 88% (proportion with insurance in sample)
Problem 2 Understanding Sampling Variability Easy
A population has μ = 50. Three random samples of size 25 are taken, yielding sample means of 48.2, 51.5, and 49.8. Explain why the sample means are different from each other and from μ.
Solution:
The sample means differ due to sampling variability (or sampling error).
Explanation:
- Each sample is randomly selected, so each includes different individuals from the population
- By random chance, some samples will include more high values, others more low values
- This natural variation causes sample means to vary from sample to sample
- The sample means cluster around μ = 50 but aren't exactly 50 because each sample is a different random selection
Key point: This is normal and expected! Sampling variability is why we study sampling distributions.
Problem 3 Properties of Sampling Distribution Medium
A population has mean μ = 80 and standard deviation σ = 16. For samples of size n = 64, describe the sampling distribution of x̄ (including shape, center, and spread).
Solution:
Check CLT conditions:
- Assume random sampling
- n = 64 ≥ 30
- CLT applies!
Sampling distribution of x̄:
- Shape: Approximately normal (by CLT)
- Center: μₓ̄ = μ = 80
- Spread: σₓ̄ = σ/√n = 16/√64 = 16/8 = 2
Answer: x̄ ~ N(80, 2) — The sampling distribution is approximately normal with mean 80 and standard error 2.
Problem 4 Notation Practice Easy
Match each symbol with its meaning:
(a) μ (b) x̄ (c) σ (d) s (e) p (f) p̂
Choices: population mean, sample mean, population SD, sample SD, population proportion, sample proportion
Solution:
- (a) μ = population mean
- (b) x̄ = sample mean
- (c) σ = population standard deviation
- (d) s = sample standard deviation
- (e) p = population proportion
- (f) p̂ = sample proportion
Problem 5 Mean of Sampling Distribution Easy
True or False: The mean of the sampling distribution of x̄ always equals the population mean μ.
Solution:
TRUE
Explanation: One of the key properties of sampling distributions is that μₓ̄ = μ. This means sample means are unbiased estimators of the population mean—on average, they equal the true population value. Some samples will have x̄ > μ and others x̄ < μ, but the average of all possible sample means equals μ.
Problems 6-10: Central Limit Theorem Applications
Problem 6 Basic CLT Application Medium
A population has μ = 120 and σ = 30. For samples of size 100, find P(x̄ < 115).
Solution:
Step 1: Check CLT — n = 100 ≥ 30
Step 2: Sampling distribution
- μₓ̄ = 120
- σₓ̄ = σ/√n = 30/√100 = 30/10 = 3
Step 3: Calculate z-score
z = (x̄ - μ) / σₓ̄ = (115 - 120) / 3 = -5/3 = -1.67
Step 4: Find probability
P(x̄ < 115) = P(z < -1.67) = 0.0475
Answer: 0.0475 or 4.75% — There's about a 4.75% chance the sample mean will be less than 115.
Problem 7 Finding a Range for Sample Means Medium
Battery life has μ = 500 hours and σ = 40 hours. For samples of 64 batteries, find the range that contains the middle 95% of sample means.
Solution:
Step 1: Sampling distribution
- μₓ̄ = 500
- σₓ̄ = 40/√64 = 40/8 = 5
Step 2: Find z-scores for middle 95%
Middle 95% → z = ±1.96
Step 3: Convert to x̄ values
- Lower: x̄ = μ + z·σₓ̄ = 500 + (-1.96)(5) = 500 - 9.8 = 490.2
- Upper: x̄ = μ + z·σₓ̄ = 500 + (1.96)(5) = 500 + 9.8 = 509.8
Answer: Between 490.2 and 509.8 hours — 95% of sample means fall in this range.
Problem 8 Probability Between Two Values Hard
Daily water usage has μ = 300 gallons and σ = 50 gallons. For 36 randomly selected days, find P(290 < x̄ < 310).
Solution:
Step 1: Sampling distribution
- μₓ̄ = 300
- σₓ̄ = 50/√36 = 50/6 ≈ 8.33
Step 2: Calculate z-scores
- For x̄ = 290: z = (290 - 300) / 8.33 = -10/8.33 ≈ -1.20
- For x̄ = 310: z = (310 - 300) / 8.33 = 10/8.33 ≈ 1.20
Step 3: Find probability
- P(-1.20 < z < 1.20) = P(z < 1.20) - P(z < -1.20)
- = 0.8849 - 0.1151 = 0.7698
Answer: 0.7698 or 76.98% — About 77% of sample means fall between 290 and 310 gallons.
Problem 9 Individual vs. Sample Mean Hard
Weights of packages have μ = 10 lbs and σ = 2 lbs. Compare:
(a) P(one package weighs more than 12 lbs)
(b) P(mean weight of 25 packages exceeds 12 lbs)
Solution:
(a) Individual package:
- z = (12 - 10) / 2 = 1.0
- P(x > 12) = P(z > 1.0) = 1 - 0.8413 = 0.1587
Answer (a): 15.87% — About 16% chance for one package
(b) Sample mean of 25:
- σₓ̄ = 2/√25 = 2/5 = 0.4
- z = (12 - 10) / 0.4 = 2/0.4 = 5.0
- P(x̄ > 12) = P(z > 5.0) ≈ 0.0000003 (essentially 0)
Answer (b): ≈ 0% — Virtually no chance for the mean of 25 packages
Key insight: Individual values vary much more than sample means! Sample means are much more predictable.
Problem 10 When CLT Doesn't Apply Medium
For which scenarios is the Central Limit Theorem NOT appropriate?
(a) Population is normally distributed, n = 20
(b) Population is strongly skewed, n = 15
(c) Population is uniform, n = 50
Solution:
(a) Normal population, n = 20: CLT DOES apply — When population is normal, any sample size works
(b) Strongly skewed, n = 15: CLT does NOT apply — n < 30 and population not normal
(c) Uniform, n = 50: CLT DOES apply — n ≥ 30 (even though population not normal)
Answer: Only (b) is inappropriate — Need larger sample for skewed populations.
Problems 11-15: Sampling Distribution of Proportions
Problem 11 Calculating Sample Proportion Easy
In a survey of 500 people, 320 prefer brand A. Calculate the sample proportion and verify normal approximation is appropriate.
Solution:
Sample proportion:
p̂ = x/n = 320/500 = 0.64 or 64%
Check success-failure condition:
- np = 500(0.64) = 320 ≥ 10
- n(1-p) = 500(0.36) = 180 ≥ 10
Answer: p̂ = 0.64, and normal approximation IS appropriate.
Problem 12 Standard Error for Proportions Medium
If p = 0.4 and n = 200, calculate the standard error of p̂.
Solution:
σₚ̂ = √(p(1-p)/n)
σₚ̂ = √(0.4 × 0.6 / 200)
σₚ̂ = √(0.24 / 200)
σₚ̂ = √0.0012
σₚ̂ ≈ 0.0346
Answer: SE ≈ 0.035 or 3.5% — Sample proportions typically vary about 3.5% from the true proportion.
Problem 13 Probability with Sample Proportions Hard
Nationally, 30% of adults exercise regularly (p = 0.30). In a sample of 150 adults, find P(p̂ > 0.35).
Solution:
Step 1: Check conditions
- np = 150(0.30) = 45 ≥ 10
- n(1-p) = 150(0.70) = 105 ≥ 10
Step 2: Find SE
σₚ̂ = √(0.30 × 0.70 / 150) = √(0.21/150) = √0.0014 ≈ 0.0374
Step 3: Calculate z-score
z = (0.35 - 0.30) / 0.0374 = 0.05 / 0.0374 ≈ 1.34
Step 4: Find probability
P(p̂ > 0.35) = P(z > 1.34) = 1 - 0.9099 = 0.0901
Answer: 0.0901 or 9.01% — About 9% chance the sample proportion exceeds 35%.
Problem 14 Testing a Claim Hard
A company claims 20% of its customers are repeat buyers. A sample of 300 customers includes 75 repeat buyers (p̂ = 0.25). Is this unusually high if the claim is true?
Solution:
Assuming p = 0.20:
Step 1: Check conditions
- np = 300(0.20) = 60 ≥ 10
- n(1-p) = 300(0.80) = 240 ≥ 10
Step 2: Find SE
σₚ̂ = √(0.20 × 0.80 / 300) = √(0.16/300) ≈ 0.0231
Step 3: Calculate z
z = (0.25 - 0.20) / 0.0231 = 0.05 / 0.0231 ≈ 2.16
Step 4: Find probability
P(p̂ ≥ 0.25) = P(z ≥ 2.16) = 1 - 0.9846 = 0.0154
Answer: YES, this is unusual. Only 1.54% chance of getting p̂ ≥ 0.25 if true p = 0.20. The company's claim may be incorrect.
Problem 15 When Normal Approximation Fails Medium
For p = 0.02 and n = 100, can we use normal approximation for the sampling distribution of p̂? Why or why not?
Solution:
Check success-failure condition:
- np = 100(0.02) = 2 < 10
- n(1-p) = 100(0.98) = 98 ≥ 10
Answer: NO, we cannot use normal approximation.
Explanation: The first condition fails (np = 2 < 10). When p is very small (or very large), we need a larger sample size for normal approximation to work. In this case, we'd need at least n ≥ 500 to get np ≥ 10.
Problems 16-20: Standard Error Calculations and Applications
Problem 16 SE vs. SD Conceptual Easy
Explain the difference between σ = 10 and SE = 2 in the context of measuring heights.
Solution:
σ = 10: Individual heights typically vary about 10 inches from the population mean. This describes the spread of the population.
SE = 2: Sample mean heights (from samples of some size n) typically vary about 2 inches from the true population mean. This describes the precision of our estimates.
Key difference: σ tells us about individual variability; SE tells us about variability of sample statistics.
Problem 17 Sample Size Effect on SE Medium
A population has σ = 60. If SE = 6 for sample size n, what sample size gives SE = 3?
Solution:
Method 1: Using the rule
To cut SE in half (from 6 to 3), we need to multiply n by 4.
First, find current n: SE = σ/√n → 6 = 60/√n → √n = 10 → n = 100
New n = 4 × 100 = 400
Method 2: Direct calculation
Want: 3 = 60/√n
3√n = 60
√n = 20
n = 400
Answer: n = 400
Problem 18 Quality Control Application Hard
Bottles should contain μ = 500ml with σ = 5ml. A quality inspector samples 25 bottles each hour. If x̄ < 497ml or x̄ > 503ml, production is stopped. What's the probability of a false alarm (stopping when process is actually fine)?
Solution:
Step 1: Find SE
σₓ̄ = 5/√25 = 5/5 = 1
Step 2: Calculate z-scores
- For 497: z = (497-500)/1 = -3.0
- For 503: z = (503-500)/1 = 3.0
Step 3: Find probabilities
- P(x̄ < 497) = P(z < -3.0) = 0.0013
- P(x̄ > 503) = P(z > 3.0) = 0.0013
- P(false alarm) = 0.0013 + 0.0013 = 0.0026
Answer: 0.0026 or 0.26% — Very low false alarm rate (only about 1 in 385 samples).
Problem 19 Sample Size for Desired Precision Hard
A researcher wants to estimate mean income with SE ≤ $500. If σ = $8000, what minimum sample size is needed?
Solution:
Want: SE ≤ 500
σ/√n ≤ 500
8000/√n ≤ 500
8000 ≤ 500√n
16 ≤ √n
256 ≤ n
Answer: n ≥ 256 — Need at least 256 people in the sample.
Problem 20 Comprehensive Application Hard
A standardized test has μ = 100 and σ = 15. A school's 36 students have x̄ = 106.
(a) Find the probability of getting x̄ ≥ 106 if the school's students are typical.
(b) Based on this probability, does the school appear to perform better than average?
Solution:
(a) Finding the probability:
Step 1: Sampling distribution
- μₓ̄ = 100
- σₓ̄ = 15/√36 = 15/6 = 2.5
Step 2: Calculate z
z = (106 - 100) / 2.5 = 6/2.5 = 2.4
Step 3: Find probability
P(x̄ ≥ 106) = P(z ≥ 2.4) = 1 - 0.9918 = 0.0082
Answer (a): 0.0082 or 0.82%
(b) Interpretation:
YES, the school appears to perform better than average. If the school were typical (μ = 100), there would be less than a 1% chance of getting a sample mean as high as 106. This low probability suggests the school's true mean is likely higher than the population mean—evidence of better-than-average performance.