Free Response Practice

Comparing Distributions

Unit 1 Exploring Data

A teacher recorded the number of absences for students in two sections of AP Statistics during a semester:

	Section A (n=28)	Section B (n=32)
Mean	3.5	5.8
Median	3.0	4.0
SD	2.1	4.3
Min	0	0
Q1	2	2
Q3	5	8
Max	9	19

(a) Compare the distributions of absences for the two sections. Include shape, center, spread, and unusual features.

(b) The teacher wants to report a "typical" number of absences for each section. Should the teacher use the mean or the median for Section B? Justify your answer.

Your Answer:

Part (a): Compare Distributions (E/P/I)

Essentially Correct (E)

Compares BOTH distributions on shape, center, AND spread using comparative language.

Partially Correct (P)

Describes each separately without comparison, OR compares only 1-2 of the three elements.

Incorrect (I)

Only lists numbers without comparison or interpretation.

Sample E Response:
Both distributions appear skewed right since the mean exceeds the median in both sections. Section B has a higher center than Section A (median 4 vs. 3, mean 5.8 vs. 3.5). Section B also has more variability (SD 4.3 vs. 2.1, IQR of 6 vs. 3). Section B has a potential outlier at 19 (upper fence = Q3 + 1.5×IQR = 8 + 9 = 17, and 19 > 17).

Part (b): Mean vs. Median (E/P/I)

Essentially Correct (E)

Chooses median AND explains that Section B is skewed right / has potential outliers, so the median is more resistant and better represents a "typical" value.

Partially Correct (P)

Chooses median but gives weak justification.

Sample E Response:
The teacher should use the median. Section B is skewed right (mean 5.8 > median 4.0) and may have an outlier at 19, so the median better represents the typical student because it is resistant to the influence of outliers and skewness.

Score Yourself:

Part (a):

Part (b):

Normal Distributions and z-Scores

Unit 1Normal Models

The scores on a standardized math test are approximately Normally distributed with a mean of 500 and a standard deviation of 100.

(a) What proportion of students score above 650? Show your work.

(b) A student scored at the 90th percentile. What was their approximate score? Show your work.

(c) A different test has a mean of 24 and SD of 4. A student scored 30 on this test and 680 on the first test. On which test did the student perform better relative to other test-takers? Justify using z-scores.

Your Answer:

Part (a): Proportion above 650 (E/P/I)

Essentially Correct (E)

z = (650-500)/100 = 1.5. P(Z > 1.5) = 1 - 0.9332 = 0.0668 (about 6.68%). Shows z-score computation and uses table or calculator.

Part (b): 90th Percentile (E/P/I)

Essentially Correct (E)

Uses invNorm(0.90, 500, 100) or finds z = 1.28, then x = 500 + 1.28(100) = 628.

Part (c): Comparing z-scores (E/P/I)

Essentially Correct (E)

Test 1: z = (680-500)/100 = 1.80. Test 2: z = (30-24)/4 = 1.50. The student performed relatively better on Test 1 because the z-score (1.80) is higher, meaning they were further above the mean in standard deviation units.

Score Yourself:

Part (a):

Part (b):

Part (c):

Interpreting a Regression

Unit 2Regression

A researcher studies the relationship between hours of sleep (x) and reaction time in milliseconds (y) for 20 adults. The LSRL is: y-hat = 450 - 15.2x, with r = -0.87 and r² = 0.757.

(a) Interpret the slope in context.

(b) Interpret r² in context.

(c) One subject slept 7 hours and had a reaction time of 360 ms. Calculate and interpret the residual.

(d) Would it be appropriate to use this model to predict reaction time for someone who slept 2 hours? Explain.

Your Answer:

Part (a): Slope (E/P/I)

Essentially Correct (E)

For each additional hour of sleep, the predicted reaction time decreases by 15.2 milliseconds. Must include "predicted" and context.

Part (b): r² (E/P/I)

Essentially Correct (E)

75.7% of the variability in reaction time is explained by the linear relationship with hours of sleep.

Part (c): Residual (E/P/I)

Essentially Correct (E)

Predicted: y-hat = 450 - 15.2(7) = 343.6. Residual = 360 - 343.6 = 16.4 ms. This person's actual reaction time was 16.4 ms slower than the model predicted.

Part (d): Extrapolation (E/P/I)

Essentially Correct (E)

No, if 2 hours is outside the range of the observed sleep data, this would be extrapolation and the linear pattern may not hold at extreme values.

Score Yourself:

Part (a):

Part (b):

Part (c):

Part (d):

Residual Plots and Model Assessment

Unit 2Model Fit

A student fits a linear model to data relating the age of a used car (in years) to its selling price (in thousands of dollars). The LSRL is: price-hat = 28.5 - 2.4(age). The residual plot shows a clear curved (U-shaped) pattern.

(a) Based on the residual plot, is a linear model appropriate for these data? Explain.

(b) The student observes that r² = 0.89. Does this high r² value mean the linear model is a good fit? Explain.

Your Answer:

Part (a): Residual Plot (E/P/I)

Essentially Correct (E)

No, the linear model is not appropriate. A pattern (curved shape) in the residual plot indicates the linear model does not adequately capture the relationship. A good residual plot should show random scatter with no pattern.

Part (b): High r² (E/P/I)

Essentially Correct (E)

No. A high r² alone does not mean the model is a good fit. The residual plot is more informative about model appropriateness than r². It is possible to have a high r² with a model that systematically over- and under-predicts in a pattern, which is what the curved residual plot reveals.

Score Yourself:

Part (a):

Part (b):

Designing an Experiment

Unit 3Study Design

A physical therapist wants to compare the effectiveness of two stretching routines (Routine A and Routine B) on improving flexibility. She has 40 volunteers.

(a) Describe a completely randomized design for this experiment. Be specific about how you would assign subjects and what you would measure.

(b) Explain why random assignment is important in this experiment.

(c) Describe how she could redesign this as a matched pairs experiment. What is one advantage?

Your Answer:

Part (a): CRD (E/P/I)

Essentially Correct (E)

(1) Randomly assign 40 volunteers to two groups of 20, (2) one does Routine A and the other Routine B, (3) compare flexibility measurements.

Sample E: Randomly assign the 40 volunteers to two groups of 20 using a random number generator. Group 1 performs Routine A and Group 2 performs Routine B for the same duration. After the treatment period, measure each participant's flexibility using the sit-and-reach test and compare the mean flexibility scores.

Part (b): Why Random Assignment (E/P/I)

Essentially Correct (E)

Creates roughly equivalent groups on confounding variables (age, fitness, baseline flexibility) so differences can be attributed to the treatment.

Part (c): Matched Pairs (E/P/I)

Essentially Correct (E)

Each subject does BOTH routines in random order with a washout period. Advantage: controls for individual variation, reducing variability.

Score Yourself:

Part (a):

Part (b):

Part (c):

Sampling Bias and Study Design

Unit 3Sampling

A city council wants to know whether residents support building a new community center. They place a survey at the front desk of the current community center and collect 200 responses. 78% support the project.

(a) Identify a source of bias in this sampling method and explain the likely direction of the bias.

(b) Describe a better sampling method the city council could use to get a more representative result.

Your Answer:

Part (a): Identify Bias (E/P/I)

Essentially Correct (E)

This is a convenience / voluntary response sample. People who visit the current community center are more likely to value community facilities and support a new one. This would overestimate the true level of support among all city residents.

Part (b): Better Method (E/P/I)

Essentially Correct (E)

Take a simple random sample of city residents (e.g., randomly select 200 addresses from city records and mail or call each one). This ensures every resident has an equal chance of being selected, reducing bias.

Score Yourself:

Part (a):

Part (b):

Probability and Random Variables

Unit 4Probability

A carnival game costs $2 to play. A player spins a wheel with these outcomes:

Prize	$0	$1	$5	$20
Probability	0.50	0.30	0.15	0.05

(a) Find the expected value of the net profit per game. Show your work.

(b) Find the standard deviation of the net profit.

(c) A player plays 50 independent games. Find the expected total net profit and the standard deviation of the total.

Your Answer:

Part (a): Expected Net Profit (E/P/I)

Essentially Correct (E)

E(prize) = 0(0.50)+1(0.30)+5(0.15)+20(0.05) = $2.05. Net profit = $2.05 - $2.00 = $0.05.

Part (b): Standard Deviation (E/P/I)

Essentially Correct (E)

Use net profit values (-2, -1, 3, 18). Var = Σ(xi-μ)²·P(xi). SD ≈ $4.46. Note: subtracting the $2 cost shifts values but does NOT change SD.

Part (c): 50 Games (E/P/I)

Essentially Correct (E)

Since independent: E(total) = 50 × $0.05 = $2.50. SD(total) = $4.46 × √50 ≈ $31.54.

Score Yourself:

Part (a):

Part (b):

Part (c):

Binomial Probability

Unit 4Binomial

A manufacturer claims that 90% of its products pass quality inspection. A store receives a shipment of 20 items and inspects all of them.

(a) Explain why this situation can be modeled with a binomial distribution. Identify n and p.

(b) What is the probability that exactly 18 of the 20 items pass inspection?

(c) If only 15 of the 20 items pass, should the store doubt the manufacturer's claim? Justify using probability.

Your Answer:

Part (a): Binomial Conditions (E/P/I)

Essentially Correct (E)

Binary: pass or fail. Independent: each item's quality is independent. Fixed n = 20. Same p = 0.90 for each item.

Part (b): P(X = 18) (E/P/I)

Essentially Correct (E)

P(X=18) = C(20,18)(0.9)^18(0.1)^2 = 190(0.1501)(0.01) ≈ 0.2852.

Part (c): Doubt the Claim? (E/P/I)

Essentially Correct (E)

If p = 0.90, then P(X ≤ 15) is very small (≈ 0.0432). Since this probability is low (less than 5%), getting only 15 passes would be unusual if the claim is true, so the store has reason to doubt the 90% claim.

Score Yourself:

Part (a):

Part (b):

Part (c):

Sampling Distributions

Unit 5Sampling Distributions

A large university reports that the mean GPA of all students is 3.1 with a standard deviation of 0.6. A random sample of 36 students is selected.

(a) Describe the sampling distribution of the sample mean GPA. Include shape, mean, and standard deviation. Justify any conditions.

(b) What is the probability that the sample mean GPA is greater than 3.3?

(c) Would your answer to part (a) change if the sample size were 10 instead of 36? Explain.

Your Answer:

Part (a): Describe Sampling Distribution (E/P/I)

Essentially Correct (E)

Shape: approximately Normal by CLT since n = 36 ≥ 30. Mean: μ(x-bar) = 3.1. SD: σ/√n = 0.6/√36 = 0.1.

Part (b): Probability (E/P/I)

Essentially Correct (E)

z = (3.3 - 3.1)/0.1 = 2.0. P(x-bar > 3.3) = P(Z > 2) = 0.0228.

Part (c): Smaller Sample (E/P/I)

Essentially Correct (E)

Yes. With n = 10, we can't rely on CLT (n < 30). We would need the population distribution of GPAs to be approximately Normal to claim the sampling distribution is Normal. The SD would also increase to 0.6/√10 ≈ 0.19.

Score Yourself:

Part (a):

Part (b):

Part (c):

Sampling Distribution of a Proportion

Unit 5Proportions

Suppose 35% of all adults in a city exercise regularly. A random sample of 150 adults is selected.

(a) Check the conditions for the sampling distribution of p-hat to be approximately Normal.

(b) Find the mean and standard deviation of the sampling distribution of p-hat.

(c) What is the probability that more than 40% of the sample exercises regularly?

Your Answer:

Part (a): Conditions (E/P/I)

Essentially Correct (E)

np = 150(0.35) = 52.5 ≥ 10 and n(1-p) = 150(0.65) = 97.5 ≥ 10. Both met, so the sampling distribution of p-hat is approximately Normal.

Part (b): Mean and SD (E/P/I)

Essentially Correct (E)

Mean = p = 0.35. SD = √(0.35×0.65/150) = √(0.001517) ≈ 0.0389.

Part (c): P(p-hat > 0.40) (E/P/I)

Essentially Correct (E)

z = (0.40-0.35)/0.0389 = 1.285. P(Z > 1.285) ≈ 0.0994 (about 10%).

Score Yourself:

Part (a):

Part (b):

Part (c):

Significance Test for a Proportion

Unit 6Inference

A polling organization claims 60% of voters support a new park. A reporter suspects the proportion is different and surveys 150 random voters, finding 80 supporters.

(a) Conduct a significance test at α = 0.05 using the full 4-step inference procedure.

(b) Based on your test, would a 95% confidence interval contain 0.60? Explain without calculating.

Your Answer:

Part (a): Full 4-Step Procedure

STATE (E)

Let p = true proportion of voters who support the park. H0: p = 0.60, Ha: p ≠ 0.60. α = 0.05.

PLAN (E)

One-proportion z-test. Random: stated random sample. 10%: 150 ≤ 10% of all voters. Large counts: 150(0.60)=90≥10, 150(0.40)=60≥10.

DO (E)

p-hat = 80/150 = 0.5333. z = (0.5333-0.60)/√(0.60×0.40/150) = -0.0667/0.0400 = -1.67. p-value = 2P(Z<-1.67) ≈ 0.095.

CONCLUDE (E)

Since p-value (0.095) > α (0.05), we fail to reject H0. There is not convincing evidence that the true proportion differs from 0.60.

Part (b): CI Connection (E/P/I)

Essentially Correct (E)

Yes, 0.60 would be in the 95% CI. Since we failed to reject at α=0.05, 0.60 is a plausible value and would fall within the interval.

Score Yourself:

State:

Plan:

Do:

Conclude:

Part (b):

Confidence Interval for a Proportion

Unit 6Confidence Interval

In a random sample of 500 high school seniors, 215 say they plan to attend a 4-year university.

(a) Construct a 95% confidence interval for the proportion of all seniors who plan to attend a 4-year university. Follow the full inference procedure.

(b) Interpret the interval in context.

(c) A guidance counselor claims that a majority of seniors plan to attend a 4-year university. Does your interval support this claim? Explain.

Your Answer:

Part (a): Construct CI (E/P/I)

Essentially Correct (E)

p-hat = 215/500 = 0.43. Conditions: random sample, 500 ≤ 10% of all seniors, 500(0.43)=215≥10 and 500(0.57)=285≥10. CI: 0.43 ± 1.96√(0.43×0.57/500) = 0.43 ± 0.0434 = (0.387, 0.473).

Part (b): Interpret (E/P/I)

Essentially Correct (E)

We are 95% confident that the true proportion of all high school seniors who plan to attend a 4-year university is between 0.387 and 0.473.

Part (c): Majority? (E/P/I)

Essentially Correct (E)

No. The entire interval (0.387 to 0.473) is below 0.50, so the data do not support the claim that a majority plan to attend.

Score Yourself:

Part (a):

Part (b):

Part (c):

One-Sample t-Test for a Mean

Unit 7Inference

A cereal company claims each box contains 16 oz. A consumer group suspects the boxes are underfilled. They randomly select 25 boxes and find x-bar = 15.7 oz with s = 0.8 oz.

(a) Conduct a significance test at α = 0.05. Use the full 4-step procedure.

(b) Based on your conclusion, what type of error (Type I or Type II) could you have made? Describe it in context.

Your Answer:

Part (a): Full 4-Step Procedure

STATE (E)

μ = true mean weight of cereal boxes. H0: μ = 16, Ha: μ < 16 (one-sided, suspects underfilling). α = 0.05.

PLAN (E)

One-sample t-test. Random: stated random sample of 25. 10%: 25 ≤ 10% of all boxes. Normal: n=25 is close to 30; assume no strong skewness (or state we'd need to check a graph).

DO (E)

t = (15.7-16)/(0.8/√25) = -0.3/0.16 = -1.875, df = 24. p-value = P(t < -1.875) ≈ 0.0365.

CONCLUDE (E)

Since p-value (0.0365) < α (0.05), we reject H0. There is convincing evidence that the true mean weight of cereal boxes is less than 16 oz.

Part (b): Type of Error (E/P/I)

Essentially Correct (E)

We rejected H0, so we could have made a Type I error — concluding that the boxes are underfilled when in reality the true mean weight is 16 oz.

Score Yourself:

State:

Plan:

Do:

Conclude:

Part (b):

Two-Sample t-Test

Unit 7Two-Sample

A teacher wants to compare test performance between two class sections. Section 1 (n=30): x-bar=78.2, s=10.5. Section 2 (n=28): x-bar=82.8, s=9.3.

(a) Is there convincing evidence of a difference in mean test scores? Conduct a two-sample t-test at α = 0.05.

(b) A 95% confidence interval for μ1 - μ2 is (-9.88, 0.68). Interpret this interval and explain how it's consistent with your test result.

Your Answer:

Part (a): Two-Sample t-test (E/P/I per step)

Sample Response:
STATE: μ1 = true mean score for Section 1, μ2 = true mean for Section 2. H0: μ1 = μ2. Ha: μ1 ≠ μ2. α=0.05.
PLAN: Two-sample t-test. Both random/representative, independent groups, both n≥28 so CLT applies.
DO: t = (78.2-82.8)/√(10.5²/30 + 9.3²/28) = -4.6/2.61 ≈ -1.76. df ≈ 55 (calculator). p-value ≈ 0.084.
CONCLUDE: p-value (0.084) > α (0.05), fail to reject H0. No convincing evidence of a difference.

Part (b): CI Interpretation (E/P/I)

Essentially Correct (E)

We are 95% confident the true difference (μ1 - μ2) is between -9.88 and 0.68. Since 0 is in the interval, we cannot conclude a significant difference — consistent with failing to reject H0.

Score Yourself:

Part (a):

Part (b):

Chi-Square Test for Independence

Unit 8Chi-Square

400 adults were surveyed about news source preference and age group:

	TV	Online	Print	Total
18-39	30	110	10	150
40-64	55	70	25	150
65+	65	20	15	100
Total	150	200	50	400

(a) State appropriate hypotheses.

(b) Calculate the expected count for "18-39, TV." Show the formula.

(c) χ² = 82.44, p-value ≈ 0. State your conclusion at α = 0.05.

(d) Which cells contribute most to χ²? What does this reveal about the association?

Your Answer:

Part (a): Hypotheses (E/P/I)

Essentially Correct (E)

H0: News source preference and age group are independent. Ha: They are not independent (there IS an association).

Part (b): Expected Count (E/P/I)

Essentially Correct (E)

Expected = (row total × column total)/grand total = (150×150)/400 = 56.25.

Part (c): Conclusion (E/P/I)

Essentially Correct (E)

p-value ≈ 0 < 0.05, so reject H0. Convincing evidence that news source preference and age are associated.

Part (d): Follow-up (E/P/I)

Essentially Correct (E)

"18-39, Online" (observed 110, expected 75) and "65+, Online" (observed 20, expected 50) contribute most. Younger adults strongly prefer online; older adults use it far less than expected.

Score Yourself:

Part (a):

Part (b):

Part (c):

Part (d):

Chi-Square Goodness of Fit

Unit 8GOF

A die manufacturer claims their die is fair. You roll it 300 times and get: 1 (58), 2 (45), 3 (52), 4 (49), 5 (51), 6 (45).

(a) State hypotheses for a goodness-of-fit test.

(b) What is the expected count for each face? Check the Large Counts condition.

(c) Calculate the chi-square statistic and state the degrees of freedom.

Your Answer:

Part (a): Hypotheses (E/P/I)

Essentially Correct (E)

H0: The die is fair (each face has probability 1/6). Ha: The die is not fair (at least one probability differs from 1/6).

Part (b): Expected Counts (E/P/I)

Essentially Correct (E)

Expected = 300/6 = 50 for each face. All expected counts = 50 ≥ 5, so Large Counts condition is met.

Part (c): Chi-Square Statistic (E/P/I)

Essentially Correct (E)

χ² = (58-50)²/50 + (45-50)²/50 + (52-50)²/50 + (49-50)²/50 + (51-50)²/50 + (45-50)²/50 = 1.28+0.50+0.08+0.02+0.02+0.50 = 2.40. df = 6-1 = 5.

Score Yourself:

Part (a):

Part (b):

Part (c):

Regression Inference: t-Test for Slope

Unit 9Investigative Task

A biologist studies temperature (°F) and cricket chirps per minute at 15 locations. Computer output:

Predictor	Coef	SE Coef	T	P
Constant	-0.31	3.10	-0.10	0.922
Temperature	0.212	0.0437	4.85	0.0003

S = 3.58 R-sq = 64.4%

(a) Write the LSRL equation. Define variables.

(b) Interpret the slope in context.

(c) Interpret r² in context.

(d) Is there convincing evidence of a linear relationship? Conduct a test at α = 0.05 using the output.

(e) Construct a 95% CI for the true slope. (t* = 2.160, df = 13)

Your Answer:

Part (a): Equation (E/P/I)

Essentially Correct (E)

chirps-hat = -0.31 + 0.212(temperature). Must use y-hat notation and define variables.

Part (b): Slope (E/P/I)

Essentially Correct (E)

For each additional °F, the predicted number of chirps per minute increases by 0.212.

Part (c): r² (E/P/I)

Essentially Correct (E)

64.4% of the variability in chirps per minute is explained by the linear relationship with temperature.

Part (d): t-test (E/P/I)

Sample: H0: β=0, Ha: β≠0, α=0.05. t-test for slope. From output: t=4.85, p=0.0003. Since 0.0003 < 0.05, reject H0. Convincing evidence of a linear relationship.

Part (e): CI for Slope (E/P/I)

Essentially Correct (E)

0.212 ± 2.160(0.0437) = 0.212 ± 0.0944 = (0.118, 0.306). We are 95% confident the true slope is between 0.118 and 0.306 chirps per °F.

Score Yourself:

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Part (b):

Part (c):

Part (d):

Part (e):

Checking LINER Conditions

Unit 9Conditions

A student fits a regression to predict house price from square footage (n = 40). The scatterplot looks roughly linear. The residual plot shows a slight fan shape (spread increases with x). A histogram of residuals is approximately symmetric and bell-shaped.

(a) Check each of the five LINER conditions using the information provided. For each, state whether it is met, not met, or cannot be determined.

(b) The "Equal variance" condition appears violated. What impact might this have on inference for the slope?

Your Answer:

Part (a): LINER Conditions (E/P/I)

Essentially Correct (E)

Linear: Met — scatterplot looks roughly linear.
Independent: Cannot fully determine from given info, but n=40 likely < 10% of all houses.
Normal: Met — histogram of residuals is approximately symmetric and bell-shaped.
Equal variance: NOT met — fan shape in residual plot means spread is not constant.
Random: Cannot be determined — not stated whether houses were randomly selected.

Part (b): Impact of Violated Condition (E/P/I)

Essentially Correct (E)

If equal variance is violated, the standard error of the slope may be inaccurate, which means the confidence interval and p-value for the slope test may not be reliable. The test could be too liberal or too conservative depending on the pattern.

Score Yourself:

Part (a):

Part (b):

Describing a Distribution from a Graph

Unit 1SOCS

A histogram of 200 students' commute times (in minutes) to school shows a unimodal shape with a long tail to the right. The mean is 22 minutes, the median is 18 minutes, the standard deviation is 12 minutes, and the IQR is 14 minutes. One student has a commute of 65 minutes.

(a) Describe the distribution of commute times. Address all four components (shape, outliers, center, spread).

(b) Is the student with a 65-minute commute an outlier by the 1.5 × IQR rule? Show your work. (Q1 = 12, Q3 = 26)

(c) Would you recommend the mean or median to describe a typical commute time? Justify.

Your Answer:

Part (a): SOCS Description (E/P/I)

Essentially Correct (E)

Shape: unimodal, skewed right. Center: median = 18 min (or mean = 22). Spread: SD = 12 or IQR = 14. Outliers: the 65-min value appears unusually high. Must address all four in context.

Part (b): Outlier Check (E/P/I)

Essentially Correct (E)

IQR = 26 - 12 = 14. Upper fence = Q3 + 1.5(IQR) = 26 + 21 = 47. Since 65 > 47, the 65-minute commute IS an outlier.

Part (c): Mean vs Median (E/P/I)

Essentially Correct (E)

Median. The distribution is skewed right with outliers, so the median is more resistant and better represents a typical commute.

Score Yourself:

Part (a):

Part (b):

Part (c):

Normal Distribution Calculations

Unit 1Normal Model

Battery life of a certain phone model is approximately Normally distributed with mean 11.5 hours and standard deviation 1.8 hours.

(a) What proportion of phones have a battery life between 10 and 14 hours?

(b) The company wants to advertise a guaranteed minimum battery life such that 90% of phones exceed it. What value should they use?

Your Answer:

Part (a): Proportion between 10 and 14 (E/P/I)

Essentially Correct (E)

z1 = (10-11.5)/1.8 = -0.833. z2 = (14-11.5)/1.8 = 1.389. P(-0.833 < Z < 1.389) = normalcdf(10, 14, 11.5, 1.8) ≈ 0.7154 (about 71.5%).

Part (b): 90th percentile from below = 10th percentile (E/P/I)

Essentially Correct (E)

Need P(X > value) = 0.90, so P(X < value) = 0.10. invNorm(0.10, 11.5, 1.8) or z = -1.282, x = 11.5 + (-1.282)(1.8) ≈ 9.19 hours.

Score Yourself:

Part (a):

Part (b):

Comparing Two Distributions with Boxplots

Unit 1Comparison

Side-by-side boxplots show the test scores for two classes. Class A: min=55, Q1=68, med=76, Q3=84, max=95. Class B: min=40, Q1=60, med=72, Q3=88, max=100.

(a) Compare the two distributions in context (center, spread, shape).

(b) Which class performed more consistently? Justify using a specific measure of spread.

Your Answer:

Part (a): Compare (E/P/I)

Essentially Correct (E)

Center: Class A has a higher median (76 vs 72). Spread: Class B has more variability (IQR = 28 vs 16; range = 60 vs 40). Shape: Both appear roughly symmetric (median near center of box). Must use comparative language.

Part (b): Consistency (E/P/I)

Essentially Correct (E)

Class A performed more consistently. IQR for A = 16 vs IQR for B = 28. The smaller IQR means scores in Class A were more tightly clustered.

Score Yourself:

Part (a):

Part (b):

Effect of Transformations on Summary Statistics

Unit 1Transformations

A set of test scores has mean 72 and standard deviation 8. The teacher applies a curve: new score = 1.1(old score) + 5.

(a) Find the new mean and new standard deviation.

(b) A student originally scored 80. What is their new score, and did their z-score change? Explain.

Your Answer:

Part (a): New Mean and SD (E/P/I)

Essentially Correct (E)

New mean = 1.1(72) + 5 = 79.2 + 5 = 84.2. New SD = 1.1(8) = 8.8. Adding 5 doesn't change SD.

Part (b): z-score (E/P/I)

Essentially Correct (E)

New score = 1.1(80) + 5 = 93. Old z = (80-72)/8 = 1.0. New z = (93-84.2)/8.8 = 1.0. The z-score did not change because the linear transformation shifted and stretched all values equally, preserving relative standing.

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Part (b):

Full Regression Analysis with Context

Unit 2Regression

A study of 30 cities examines the relationship between average temperature (°F) and monthly electricity usage (kWh). The LSRL is: usage-hat = 1450 - 12.3(temperature). r = -0.78, r² = 0.608.

(a) Identify the explanatory and response variables.

(b) Interpret the slope and y-intercept in context. Is the y-intercept meaningful?

(c) Interpret r² in context.

(d) Predict usage for a city with avg temp 70°F. Then explain why predicting for a city with avg temp 120°F would be problematic.

Your Answer:

Part (a): Variables (E/P/I)

Essentially Correct (E)

Explanatory: average temperature. Response: monthly electricity usage.

Part (b): Slope and Intercept (E/P/I)

Essentially Correct (E)

Slope: For each additional degree increase in temperature, predicted electricity usage decreases by 12.3 kWh. Intercept: When temperature is 0°F, predicted usage is 1450 kWh. The intercept may not be meaningful if 0°F is outside the data range.

Part (c): r² (E/P/I)

Essentially Correct (E)

60.8% of the variability in monthly electricity usage is explained by the linear relationship with average temperature.

Part (d): Prediction and Extrapolation (E/P/I)

Essentially Correct (E)

At 70°F: 1450 - 12.3(70) = 589 kWh. 120°F would be extrapolation — far beyond the range of observed temperatures — and the linear pattern may not hold.

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Influential Points and Outliers

Unit 2Influence

A scatterplot of 20 data points shows a moderate positive linear relationship with r = 0.55. One point has a very high x-value and falls far below the regression line. When this point is removed, r increases to 0.88.

(a) Is this point influential? Explain how you know.

(b) Does this point have high leverage? Explain.

(c) Should the researcher remove this point from the analysis? Discuss considerations.

Your Answer:

Part (a): Influential? (E/P/I)

Essentially Correct (E)

Yes, it is influential because removing it substantially changes the regression results (r went from 0.55 to 0.88).

Part (b): High leverage? (E/P/I)

Essentially Correct (E)

Yes, high leverage because its x-value is far from the mean of x. Points with extreme x-values have high leverage.

Part (c): Remove it? (E/P/I)

Essentially Correct (E)

Not necessarily. First investigate WHY it's unusual — is it a data entry error (remove it), a legitimately different case (report results with and without it), or something else? Removing data simply to improve r is not good practice without justification.

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Scope of Inference

Unit 3Generalization & Causation

Three studies examine the relationship between sleep and academic performance:

Study 1: A random sample of 500 college students is surveyed about sleep hours and GPA. More sleep is associated with higher GPA.

Study 2: 80 volunteers are randomly assigned to either a "sleep coaching" program or a control group. After 8 weeks, the coached group has significantly higher GPAs.

Study 3: Researchers randomly select 200 students from a university and randomly assign them to sleep coaching or control. The coached group shows higher GPAs.

(a) For each study, state whether we can generalize to a larger population and whether we can conclude causation. Justify each.

Your Answer:

Part (a): Scope of Inference for Each Study (E/P/I)

Essentially Correct (E)

Study 1: Random selection → CAN generalize to all college students. No random assignment → CANNOT conclude causation (observational study).
Study 2: Volunteers, not random selection → CANNOT generalize. Random assignment → CAN conclude causation (for these volunteers).
Study 3: Random selection → CAN generalize. Random assignment → CAN conclude causation. This is the strongest design.

Score Yourself:

Part (a):

Describing an Experimental Design

Unit 3Experiment

A farmer wants to test whether a new fertilizer increases crop yield compared to the standard fertilizer. He has 30 identical plots of land available. He suspects that plots on the east side of the farm receive more sunlight than those on the west side.

(a) Describe a completely randomized design.

(b) Describe a randomized block design that accounts for the sunlight difference. Explain why blocking improves the experiment.

(c) What is the purpose of having a control group (standard fertilizer) rather than comparing the new fertilizer to no fertilizer at all?

Your Answer:

Part (a): CRD (E/P/I)

Essentially Correct (E)

Randomly assign 15 plots to new fertilizer and 15 to standard. Measure and compare yields.

Part (b): Blocked Design (E/P/I)

Essentially Correct (E)

Block by location: 15 east plots and 15 west plots. Within each block, randomly assign plots to new or standard fertilizer. Blocking reduces variability due to sunlight differences, making it easier to detect a real fertilizer effect.

Part (c): Control Group (E/P/I)

Essentially Correct (E)

Using the standard fertilizer as control isolates the effect of the type of fertilizer rather than confounding it with the effect of having any fertilizer vs none.

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Conditional Probability with Two-Way Tables

Unit 4Conditional Probability

A school surveyed 400 students about transportation and grade level:

	Car	Bus	Walk	Total
Underclass (9-10)	40	120	40	200
Upperclass (11-12)	100	60	40	200
Total	140	180	80	400

(a) What is P(Car | Upperclass)?

(b) What is P(Upperclass | Car)?

(c) Are the events "Car" and "Upperclass" independent? Justify using probabilities.

Your Answer:

Part (a): P(Car | Upperclass) (E/P/I)

Essentially Correct (E)

P(Car | Upperclass) = 100/200 = 0.50.

Part (b): P(Upperclass | Car) (E/P/I)

Essentially Correct (E)

P(Upperclass | Car) = 100/140 = 0.714. Note this is different from part (a)!

Part (c): Independence (E/P/I)

Essentially Correct (E)

P(Car) = 140/400 = 0.35. P(Car | Upperclass) = 0.50. Since 0.35 ≠ 0.50, the events are NOT independent. Knowing a student is upperclass changes the probability of driving.

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Combining Random Variables

Unit 4Random Variables

A coffee shop sells small coffees (mean profit $1.50, SD $0.40) and pastries (mean profit $2.00, SD $0.60). Assume the profit from a coffee sale and a pastry sale are independent.

(a) A customer buys one coffee and one pastry. Find the mean and SD of the total profit.

(b) Another customer buys 2 coffees (no pastry). Find the mean and SD of the total profit from both coffees.

(c) Explain why SD(coffee + coffee) ≠ 2 × SD(coffee).

Your Answer:

Part (a): Coffee + Pastry (E/P/I)

Essentially Correct (E)

Mean = 1.50 + 2.00 = $3.50. Var = 0.40² + 0.60² = 0.16 + 0.36 = 0.52. SD = √0.52 ≈ $0.72.

Part (b): 2 Coffees (E/P/I)

Essentially Correct (E)

Mean = 1.50 + 1.50 = $3.00. Var = 0.40² + 0.40² = 0.32. SD = √0.32 ≈ $0.566.

Part (c): Why Not Double SD (E/P/I)

Essentially Correct (E)

We add variances, not standard deviations. SD(C+C) = √(Var(C) + Var(C)) = √(2·Var(C)) = SD(C)·√2, not 2·SD(C). Variability grows more slowly than linearly.

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CLT and Sampling Distribution of the Mean

Unit 5CLT

Wait times at a DMV are strongly right-skewed with μ = 45 minutes and σ = 20 minutes.

(a) Can we find P(individual wait > 60 min) using the Normal distribution? Explain.

(b) For a random sample of n = 64, describe the sampling distribution of x-bar. Justify the use of the Normal model.

(c) What is P(x-bar > 50) for a sample of 64?

Your Answer:

Part (a): Individual Wait (E/P/I)

Essentially Correct (E)

No. The population is strongly right-skewed, not Normal, so we cannot use the Normal distribution for an individual observation.

Part (b): Sampling Distribution (E/P/I)

Essentially Correct (E)

By CLT, since n = 64 ≥ 30, the sampling distribution of x-bar is approximately Normal with mean = 45 and SD = 20/√64 = 2.5.

Part (c): P(x-bar > 50) (E/P/I)

Essentially Correct (E)

z = (50 - 45)/2.5 = 2.0. P(x-bar > 50) = P(Z > 2) ≈ 0.0228.

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Part (b):

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Bias and Variability of Estimators

Unit 5Estimators

Two students each collect 100 random samples from the same population (true mean μ = 50). Student A uses samples of size 10; Student B uses samples of size 100. Both plot the sampling distributions of x-bar.

(a) What will be the same about the two sampling distributions?

(b) What will be different? Be specific.

(c) Which student's estimates will be more useful for estimating μ? Why?

Your Answer:

Part (a): Same (E/P/I)

Essentially Correct (E)

Both are centered at μ = 50 (x-bar is unbiased for any sample size).

Part (b): Different (E/P/I)

Essentially Correct (E)

Student B's distribution will be much narrower (SD = σ/√100 vs σ/√10). Student B's will also be more Normal-shaped by CLT.

Part (c): More Useful (E/P/I)

Essentially Correct (E)

Student B's. Larger samples produce less variable estimates, meaning individual sample means are more likely to be close to the true μ.

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Two-Proportion z-Test

Unit 6Two Proportions

A company tests two website designs. Design A: 120 of 500 visitors made a purchase. Design B: 155 of 500 visitors made a purchase.

(a) Is there convincing evidence that Design B has a higher purchase rate? Conduct a full significance test at α = 0.05.

(b) Construct a 95% confidence interval for p_B - p_A and interpret it.

Your Answer:

Part (a): Two-Prop z-Test (E/P/I per step)

Sample:
STATE: p_A = true purchase rate for Design A, p_B for Design B. H0: p_A = p_B. Ha: p_B > p_A. α = 0.05.
PLAN: Two-proportion z-test. Random assignment (assumed). 10% condition met. p-hat pooled = 275/1000 = 0.275. Large counts: all np and n(1-p) ≥ 10.
DO: p-hat_A = 0.24, p-hat_B = 0.31. z = (0.31-0.24)/√(0.275×0.725×(1/500+1/500)) = 0.07/0.0283 ≈ 2.47. p-value ≈ 0.0068.
CONCLUDE: Since 0.0068 < 0.05, reject H0. Convincing evidence Design B has a higher purchase rate.

Part (b): CI (E/P/I)

Essentially Correct (E)

(0.31-0.24) ± 1.96√(0.24×0.76/500 + 0.31×0.69/500) = 0.07 ± 0.054 = (0.016, 0.124). We are 95% confident Design B's purchase rate is between 1.6 and 12.4 percentage points higher than Design A's.

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Part (b):

Type I/II Errors and Power

Unit 6Errors & Power

A pharmaceutical company tests whether a new drug lowers cholesterol more than the current standard. H0: The new drug is no better than the standard. Ha: The new drug is better.

(a) Describe a Type I error in this context. What would be a consequence?

(b) Describe a Type II error in this context. What would be a consequence?

(c) The company uses α = 0.01 instead of 0.05. How does this affect the probability of each type of error?

(d) Name two ways to increase the power of this test.

Your Answer:

Part (a): Type I (E/P/I)

Essentially Correct (E)

Concluding the new drug is better when it actually isn't. Consequence: patients switch to a drug that's no better, wasting money and potentially missing the real standard treatment.

Part (b): Type II (E/P/I)

Essentially Correct (E)

Failing to detect that the new drug IS better when it actually is. Consequence: an effective drug is not adopted, and patients miss out on better treatment.

Part (c): Changing α (E/P/I)

Essentially Correct (E)

Lowering α from 0.05 to 0.01 decreases P(Type I) but increases P(Type II). Harder to reject H0 means fewer false positives but more false negatives.

Part (d): Increase Power (E/P/I)

Essentially Correct (E)

Any two of: increase sample size, increase α, increase the true effect size (if possible). Most practical: larger sample.

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Matched Pairs t-Test

Unit 7Matched Pairs

A trainer measures the resting heart rate of 12 participants before and after an 8-week exercise program. The mean difference (before - after) is d-bar = 5.2 bpm with s_d = 6.8 bpm.

(a) Is there convincing evidence that the program reduces resting heart rate? Conduct a matched pairs t-test at α = 0.05.

(b) Construct a 95% confidence interval for the true mean difference and interpret it.

Your Answer:

Part (a): Matched Pairs t-test (E/P/I per step)

Sample:
STATE: μ_d = true mean difference in heart rate (before - after). H0: μ_d = 0. Ha: μ_d > 0. α = 0.05.
PLAN: Matched pairs t-test (one-sample t on differences). Random: assume random or representative. Normal: n = 12 is small — need to assume no strong skewness in differences.
DO: t = (5.2 - 0)/(6.8/√12) = 5.2/1.963 = 2.649. df = 11. p-value ≈ 0.011.
CONCLUDE: Since 0.011 < 0.05, reject H0. Convincing evidence the program reduces resting heart rate.

Part (b): CI (E/P/I)

Essentially Correct (E)

5.2 ± 2.201(6.8/√12) = 5.2 ± 4.32 = (0.88, 9.52). We are 95% confident the true mean reduction in heart rate is between 0.88 and 9.52 bpm.

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Choosing Between Procedures

Unit 7Procedure Selection

For each scenario, identify the appropriate inference procedure and justify your choice.

(a) A nutritionist randomly selects 40 adults and measures their daily calcium intake. She wants to estimate the mean intake for all adults.

(b) A researcher compares test scores between students who took an online course (n=35) and students who took an in-person course (n=40). The groups are independent.

(c) A dentist measures cavity count for 25 patients before and after switching to a new toothpaste.

(d) A pollster wants to know whether the proportion of voters supporting a candidate exceeds 50%, based on a random sample of 600 voters where 324 expressed support.

Your Answer:

All Parts (E/P/I)

Essentially Correct (E)

(a) One-sample t-interval for a mean. One group, quantitative data, estimating μ.
(b) Two-sample t-test for difference in means. Two independent groups, quantitative response.
(c) Matched pairs t-test. Same subjects measured twice (before/after). Analyze differences.
(d) One-proportion z-test. One group, categorical data (support/don't), testing a claim about p.

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Chi-Square Test for Homogeneity

Unit 8Homogeneity

Three hospitals tracked patient satisfaction (Satisfied, Neutral, Dissatisfied):

	Satisfied	Neutral	Dissatisfied	Total
Hospital A	80	30	10	120
Hospital B	75	45	30	150
Hospital C	55	40	35	130
Total	210	115	75	400

(a) Is this a test for independence or homogeneity? Explain.

(b) State the hypotheses.

(c) Calculate the expected count for Hospital A, Satisfied.

(d) The test gives χ² = 22.1 with df = 4 and p-value < 0.001. State your conclusion and identify which hospital appears most different.

Your Answer:

Part (a): Independence vs Homogeneity (E/P/I)

Essentially Correct (E)

Homogeneity. Three separate populations (hospitals) are being compared on the distribution of one variable (satisfaction). Independence would use one sample with two variables.

Part (b): Hypotheses (E/P/I)

Essentially Correct (E)

H0: The distribution of satisfaction is the same across all three hospitals. Ha: At least one hospital has a different distribution.

Part (c): Expected Count (E/P/I)

Essentially Correct (E)

Expected = (120 × 210)/400 = 63.

Part (d): Conclusion (E/P/I)

Essentially Correct (E)

p-value < 0.001 < α, reject H0. Convincing evidence that satisfaction distributions differ across hospitals. Hospital A appears most different — much higher satisfaction than expected.

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Chi-Square GOF with Genetics

Unit 8GOF

A genetics model predicts offspring in a 9:3:3:1 ratio for four phenotypes. A researcher observes 160 offspring: 85, 35, 26, 14.

(a) State the hypotheses.

(b) Calculate the expected counts for each phenotype.

(c) Verify the conditions for the chi-square test.

(d) Calculate the chi-square statistic and state the degrees of freedom.

Your Answer:

Part (a): Hypotheses (E/P/I)

Essentially Correct (E)

H0: Offspring follow the 9:3:3:1 ratio. Ha: Offspring do not follow this ratio.

Part (b): Expected Counts (E/P/I)

Essentially Correct (E)

Total parts = 16. Expected: 160(9/16) = 90, 160(3/16) = 30, 160(3/16) = 30, 160(1/16) = 10.

Part (c): Conditions (E/P/I)

Essentially Correct (E)

Random (assume representative). All expected counts ≥ 5 (90, 30, 30, 10 — all ≥ 5). 10% condition (160 ≤ 10% of all possible offspring).

Part (d): χ² and df (E/P/I)

Essentially Correct (E)

χ² = (85-90)²/90 + (35-30)²/30 + (26-30)²/30 + (14-10)²/10 = 0.278 + 0.833 + 0.533 + 1.600 = 3.244. df = 4-1 = 3.

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Full Regression Inference from Computer Output

Unit 9Full Procedure

A study examines the relationship between study hours per week (x) and exam score (y) for n = 20 students. Computer output:

Predictor	Coef	SE Coef	T	P
Constant	52.3	5.8	9.02	<0.001
StudyHours	3.45	0.82	4.21	0.0005

S = 8.4 R-sq = 49.6%

(a) Write the LSRL and interpret the slope in context.

(b) Conduct a full 4-step test for the significance of the slope at α = 0.05.

(c) Construct a 95% CI for β. (t* = 2.101 for df = 18)

(d) Interpret S = 8.4 in context.

Your Answer:

Part (a): LSRL and Slope (E/P/I)

Essentially Correct (E)

score-hat = 52.3 + 3.45(hours). For each additional hour of study per week, the predicted exam score increases by 3.45 points.

Part (b): 4-Step Test (E/P/I)

Sample: H0: β = 0, Ha: β ≠ 0, α = 0.05. t-test for slope. LINER conditions assumed met. From output: t = 4.21, p = 0.0005. Since 0.0005 < 0.05, reject H0. Convincing evidence of a linear relationship between study hours and exam score.

Part (c): CI for β (E/P/I)

Essentially Correct (E)

3.45 ± 2.101(0.82) = 3.45 ± 1.723 = (1.73, 5.17). 95% confident the true increase in exam score per additional study hour is between 1.73 and 5.17 points.

Part (d): Interpret S (E/P/I)

Essentially Correct (E)

The typical distance between actual exam scores and predicted exam scores is about 8.4 points. This is the typical prediction error of the model.

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Two-Way Tables and Association

Unit 2Categorical

A survey of 600 adults asked about exercise frequency and smoking status:

	Smoker	Non-Smoker	Total
Exercises Regularly	40	260	300
Does Not Exercise	80	220	300
Total	120	480	600

(a) Calculate the conditional distribution of smoking status for each exercise group.

(b) Is there an association between exercise and smoking? Justify using the conditional distributions.

Your Answer:

Part (a): Conditional Distributions (E/P/I)

Essentially Correct (E)

Exercisers: 40/300 = 13.3% smokers, 260/300 = 86.7% non-smokers. Non-exercisers: 80/300 = 26.7% smokers, 220/300 = 73.3% non-smokers.

Part (b): Association (E/P/I)

Essentially Correct (E)

Yes, there is an association. The proportion of smokers is higher among non-exercisers (26.7%) than exercisers (13.3%). If the variables were independent, these proportions would be similar.

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Correlation vs Causation

Unit 2Interpretation

A study finds a strong positive correlation (r = 0.82) between the number of firefighters sent to a fire and the amount of damage caused by the fire.

(a) A reporter writes: "Sending more firefighters causes more damage." Explain why this conclusion is not valid.

(b) Identify a likely lurking variable and explain how it accounts for the observed association.

(c) What type of study would be needed to establish a causal relationship? Is such a study practical here?

Your Answer:

Part (a): Why Not Causal (E/P/I)

Essentially Correct (E)

This is an observational study. Correlation does not imply causation. There was no random assignment of firefighters to fires.

Part (b): Lurking Variable (E/P/I)

Essentially Correct (E)

The severity/size of the fire is a lurking variable. Larger fires cause more damage AND require more firefighters. The fire size drives both variables.

Part (c): Causal Study (E/P/I)

Essentially Correct (E)

A randomized experiment randomly assigning different numbers of firefighters to similar fires. This is NOT practical — you can't ethically withhold firefighters from a burning building.

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Part (b):

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Identifying Bias in Surveys

Unit 3Bias

A radio host asks listeners to call in and vote on whether the minimum wage should be raised. 73% of 2,400 callers say yes.

(a) Identify the sampling method and explain why the result may not represent all adults.

(b) Describe two specific sources of bias and the likely direction of each.

Your Answer:

Part (a): Sampling Method (E/P/I)

Essentially Correct (E)

Voluntary response sample. Only people who feel strongly enough call in. This does not represent all adults — only the radio's listeners who chose to participate.

Part (b): Two Biases (E/P/I)

Essentially Correct (E)

(1) Voluntary response bias: people with strong opinions (especially in favor) are more likely to call, likely overestimating support. (2) Undercoverage: only radio listeners are represented — adults who don't listen to this station are excluded, skewing the sample.

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Confounding in Observational Studies

Unit 3Confounding

A study finds that children who eat breakfast daily have higher test scores than those who skip breakfast.

(a) Can we conclude that eating breakfast causes higher test scores? Explain.

(b) Identify two potential confounding variables and explain how each could account for the association.

(c) Describe an experiment that could help determine whether breakfast actually causes better performance.

Your Answer:

Part (a): Causation? (E/P/I)

Essentially Correct (E)

No. This is an observational study — students were not randomly assigned to eat or skip breakfast. Confounding variables may explain the association.

Part (b): Confounders (E/P/I)

Essentially Correct (E)

(1) Socioeconomic status: wealthier families can afford breakfast AND provide more academic support. (2) Parental involvement: parents who ensure breakfast may also help with homework and encourage studying.

Part (c): Experiment (E/P/I)

Essentially Correct (E)

Randomly assign students to a breakfast group (provided breakfast) or control group (no breakfast provided). Compare test scores. Random assignment would balance confounders across groups.

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Geometric Probability

Unit 4Geometric

A basketball player makes 80% of her free throws. She shoots until she misses.

(a) What is the probability that her first miss comes on the 5th shot?

(b) What is the expected number of shots until her first miss?

(c) What is the probability she makes at least 3 shots before her first miss?

Your Answer:

Part (a): P(first miss on 5th) (E/P/I)

Essentially Correct (E)

Geometric with p(miss) = 0.20. P(X=5) = (0.80)⁴(0.20) = 0.4096 × 0.20 = 0.08192.

Part (b): Expected (E/P/I)

Essentially Correct (E)

E(X) = 1/p = 1/0.20 = 5 shots.

Part (c): At Least 3 Makes (E/P/I)

Essentially Correct (E)

At least 3 makes before first miss means first miss on shot 4 or later. P(X ≥ 4) = P(first 3 are makes) = (0.80)³ = 0.512.

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Part (b):

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Independence and Probability Rules

Unit 4Rules

At a college, 60% of students have a job, 45% are in a club, and 20% have both a job and are in a club.

(a) Find P(job OR club).

(b) Find P(club | job).

(c) Are "has a job" and "is in a club" independent events? Justify mathematically.

Your Answer:

Part (a): P(job or club) (E/P/I)

Essentially Correct (E)

P(J or C) = P(J) + P(C) - P(J and C) = 0.60 + 0.45 - 0.20 = 0.85.

Part (b): P(club | job) (E/P/I)

Essentially Correct (E)

P(C|J) = P(J and C)/P(J) = 0.20/0.60 = 0.333.

Part (c): Independent? (E/P/I)

Essentially Correct (E)

Check: P(C|J) = 0.333 vs P(C) = 0.45. Since 0.333 ≠ 0.45, the events are NOT independent. Having a job decreases the probability of being in a club.

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Parameter vs Statistic

Unit 5Vocabulary

A university claims that 70% of its graduates find employment within 6 months. A newspaper surveys a random sample of 200 recent graduates and finds that 128 (64%) found employment within 6 months.

(a) Identify the population, sample, parameter, and statistic in this context.

(b) Does the sample result prove that the university's claim is false? Explain using the concept of sampling variability.

Your Answer:

Part (a): Identify Terms (E/P/I)

Essentially Correct (E)

Population: all recent graduates. Sample: the 200 surveyed graduates. Parameter: p = 0.70 (claimed true proportion). Statistic: p-hat = 0.64 (sample proportion).

Part (b): Proof? (E/P/I)

Essentially Correct (E)

No. Even if the true proportion is 0.70, we wouldn't expect every sample to yield exactly 0.70. Sampling variability means p-hat varies from sample to sample. We would need a formal significance test to determine if 0.64 is far enough from 0.70 to be convincing evidence against the claim.

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Effect of Sample Size on Sampling Distributions

Unit 5Sample Size

A population has proportion p = 0.40.

(a) For n = 25, check whether the sampling distribution of p-hat is approximately Normal.

(b) For n = 100, find the mean and SD of the sampling distribution of p-hat and check the Normal condition.

(c) How many times larger must the sample be to cut the standard deviation of p-hat in half?

Your Answer:

Part (a): n = 25 (E/P/I)

Essentially Correct (E)

np = 25(0.40) = 10 ≥ 10. n(1-p) = 25(0.60) = 15 ≥ 10. Both met, so approximately Normal (barely).

Part (b): n = 100 (E/P/I)

Essentially Correct (E)

Mean = 0.40. SD = √(0.40×0.60/100) = √0.0024 = 0.049. np = 40 ≥ 10, n(1-p) = 60 ≥ 10. Normal condition met.

Part (c): Halving SD (E/P/I)

Essentially Correct (E)

SD ∝ 1/√n. To halve SD, need √n to double, so n must be 4 times larger (quadrupled).

Score Yourself:

Part (a):

Part (b):

Part (c):

Interpreting p-values and Conclusions

Unit 6Interpretation

A test of H0: p = 0.50 vs Ha: p > 0.50 yields p-value = 0.03.

(a) Interpret the p-value in context (assume the test is about whether a majority of residents support a policy).

(b) At α = 0.05, state the conclusion in context.

(c) A student says "the p-value of 0.03 means there is a 3% chance the null hypothesis is true." Correct this interpretation.

Your Answer:

Part (a): Interpret p-value (E/P/I)

Essentially Correct (E)

Assuming the true proportion of residents who support the policy is 0.50, there is a 0.03 probability of obtaining a sample proportion as extreme as or more extreme than the one observed.

Part (b): Conclusion (E/P/I)

Essentially Correct (E)

Since p-value (0.03) < α (0.05), reject H0. There is convincing evidence that more than 50% of residents support the policy.

Part (c): Correct the Student (E/P/I)

Essentially Correct (E)

Wrong. The p-value is NOT the probability that H0 is true. It is the probability of observing data this extreme IF H0 were true. H0 is either true or false — it doesn't have a probability in the frequentist framework.

Score Yourself:

Part (a):

Part (b):

Part (c):

Sample Size for a Desired Margin of Error

Unit 6Planning

A researcher wants to estimate the proportion of adults who support a new law with a margin of error no more than 3 percentage points at the 95% confidence level.

(a) If no prior estimate of p is available, what sample size is needed? Show your work.

(b) If a pilot study suggests p ≈ 0.30, what sample size is needed? How does this compare to part (a)?

Your Answer:

Part (a): No Prior Estimate (E/P/I)

Essentially Correct (E)

Use p* = 0.5 (maximizes sample size). ME = z*√(p*(1-p*)/n). 0.03 = 1.96√(0.25/n). n = (1.96/0.03)² × 0.25 = 4268.4 → n = 1068 (round up).

Part (b): With Pilot Estimate (E/P/I)

Essentially Correct (E)

0.03 = 1.96√(0.30×0.70/n). n = (1.96/0.03)² × 0.21 = 897 (round up). This is smaller than part (a) because p(1-p) is maximized at p = 0.5; knowing p ≈ 0.30 reduces the required sample size.

Score Yourself:

Part (a):

Part (b):

Confidence Interval for a Mean

Unit 7CI for Mean

A random sample of 36 packages from an assembly line has a mean weight of 16.2 oz with s = 0.9 oz. The target weight is 16.0 oz.

(a) Construct a 95% confidence interval for the true mean weight. Follow the inference procedure.

(b) Based on your interval, is there evidence the mean weight differs from the target of 16.0 oz?

(c) If the sample size were increased to 144 (same x-bar and s), how would the interval change?

Your Answer:

Part (a): CI (E/P/I)

Essentially Correct (E)

Conditions: random sample, 36 ≤ 10% of all packages, n = 36 ≥ 30 (CLT). CI = 16.2 ± 2.030(0.9/√36) = 16.2 ± 0.305 = (15.895, 16.505).

Part (b): Evidence? (E/P/I)

Essentially Correct (E)

16.0 IS in the interval, so we do not have evidence that the mean weight differs from 16.0 oz at the 95% level.

Part (c): Larger n (E/P/I)

Essentially Correct (E)

With n = 144, SE = 0.9/√144 = 0.075. The interval would be much narrower (about half the width), potentially excluding 16.0.

Score Yourself:

Part (a):

Part (b):

Part (c):

Significance vs Practical Importance

Unit 7Interpretation

A large study (n = 10,000) tests whether a new tutoring program raises SAT scores. H0: μ = 1060 (national avg). The sample mean is x-bar = 1063 with s = 200. The t-test gives t = 1.50, p = 0.067 at α = 0.05. But a 95% CI is (1059.1, 1066.9).

(a) State the conclusion of the hypothesis test at α = 0.05.

(b) Even if the result had been significant, would a 3-point increase on the SAT be practically meaningful? Discuss.

(c) Explain why a very large sample size can lead to statistically significant results that are not practically important.

Your Answer:

Part (a): Conclusion (E/P/I)

Essentially Correct (E)

p-value (0.067) > α (0.05). Fail to reject H0. Not convincing evidence the program raises scores above the national average.

Part (b): Practical Significance (E/P/I)

Essentially Correct (E)

A 3-point increase on a 1600-point scale is trivially small and unlikely to affect college admissions or scholarships. It has no practical importance even if statistically significant.

Part (c): Large n and Significance (E/P/I)

Essentially Correct (E)

Large samples produce very small standard errors, making it possible to detect tiny differences that are real but meaningless. Statistical significance only means the difference is unlikely due to chance — not that it matters in practice.

Score Yourself:

Part (a):

Part (b):

Part (c):

Independence vs Homogeneity

Unit 8Comparison

Study A: A random sample of 500 employees at one company is classified by department AND job satisfaction level.

Study B: Random samples of 150 employees are taken from each of three different companies. Each employee is classified by job satisfaction level.

(a) Which study uses a chi-square test for independence? Which uses homogeneity? Explain.

(b) Write appropriate null and alternative hypotheses for each study.

Your Answer:

Part (a): Which Test? (E/P/I)

Essentially Correct (E)

Study A: Independence — one sample, two variables measured on each individual. Study B: Homogeneity — separate samples from multiple populations, comparing distribution of one variable.

Part (b): Hypotheses (E/P/I)

Essentially Correct (E)

Study A: H0: Department and satisfaction are independent. Ha: They are not independent. Study B: H0: The distribution of satisfaction is the same across all three companies. Ha: At least one differs.

Score Yourself:

Part (a):

Part (b):

Full Chi-Square GOF Test

Unit 8Full Procedure

A candy company claims their bags contain 30% red, 25% blue, 20% green, 15% yellow, and 10% orange candies. A student counts 200 candies: 70 red, 42 blue, 38 green, 28 yellow, 22 orange.

(a) Conduct a full chi-square goodness of fit test at α = 0.05. Include hypotheses, conditions, test statistic, and conclusion.

Your Answer:

Full GOF Test (E/P/I per component)

Sample:
STATE: H0: The distribution matches the company's claim (30/25/20/15/10). Ha: It does not.
PLAN: Chi-square GOF test. Random (assumed representative). Expected counts: 60, 50, 40, 30, 20 — all ≥ 5.
DO: χ² = (70-60)²/60 + (42-50)²/50 + (38-40)²/40 + (28-30)²/30 + (22-20)²/20 = 1.667+1.280+0.100+0.133+0.200 = 3.380. df = 4. p-value ≈ 0.496.
CONCLUDE: p-value (0.496) > α (0.05). Fail to reject H0. No convincing evidence the distribution differs from the company's claim.

Score Yourself:

Full test:

Reading and Using Computer Output

Unit 9Computer Output

Computer output for a regression of height (inches) on shoe size for n = 30 adults:

Predictor	Coef	SE Coef	T	P
Constant	49.8	4.2	11.86	<0.001
ShoeSize	1.65	0.45	3.67	0.001

S = 2.95 R-sq = 32.5%

(a) Identify the values of b, SE(b), t, and p from the output.

(b) Interpret S = 2.95 in context.

(c) A friend says "the model only explains 32.5% of the variability, so it's useless." Respond to this claim.

Your Answer:

Part (a): Identify Values (E/P/I)

Essentially Correct (E)

b = 1.65, SE(b) = 0.45, t = 3.67, p = 0.001. All from the ShoeSize row.

Part (b): Interpret S (E/P/I)

Essentially Correct (E)

The typical distance between actual heights and predicted heights from the model is about 2.95 inches.

Part (c): Respond to Claim (E/P/I)

Essentially Correct (E)

While r² = 32.5% means most variability is unexplained, the relationship IS statistically significant (p = 0.001). A low r² doesn't make a model "useless" — it means other factors also influence height. The slope is still meaningful and the model provides useful predictions.

Score Yourself:

Part (a):

Part (b):

Part (c):

CI for Slope and Significance Connection

Unit 9CI & Test Connection

A researcher studies the relationship between advertising spending (thousands of $) and sales revenue (thousands of $) for n = 18 stores. A 95% confidence interval for the slope is (0.8, 4.2).

(a) Interpret this confidence interval in context.

(b) Based on this interval, would a two-sided test of H0: β = 0 at α = 0.05 reject the null? Explain without performing the test.

(c) Would a test of H0: β = 5 at α = 0.05 reject? Explain.

Your Answer:

Part (a): Interpret CI (E/P/I)

Essentially Correct (E)

We are 95% confident that for each additional $1,000 spent on advertising, the true increase in sales revenue is between $800 and $4,200.

Part (b): Test β = 0 (E/P/I)

Essentially Correct (E)

Yes, reject H0. Since 0 is NOT in the interval (0.8, 4.2), 0 is not a plausible value for β at the 5% level.

Part (c): Test β = 5 (E/P/I)

Essentially Correct (E)

Yes, reject H0: β = 5. Since 5 is NOT in the interval (0.8, 4.2), 5 is not a plausible value for β at the 5% level.

Score Yourself:

Part (a):

Part (b):

Part (c):

Transformations and Nonlinear Relationships

Unit 9Transformations

A biologist studies the relationship between the age (x, in days) and weight (y, in grams) of a growing organism. A scatterplot of y vs x shows a clear curved pattern. The residual plot for a linear model confirms a nonlinear relationship. The biologist then takes ln(y) and fits a linear model to ln(y) vs x. The new residual plot shows random scatter.

(a) Why was the original linear model inappropriate?

(b) The regression equation for the transformed data is: ln(y-hat) = 1.2 + 0.045x. What type of model does this suggest for the original (untransformed) data?

(c) Use the model to predict the weight of a 30-day-old organism.

Your Answer:

Part (a): Why Linear Failed (E/P/I)

Essentially Correct (E)

The residual plot for the linear model showed a pattern (curved), indicating the linear model did not capture the true relationship. A good model produces residuals with no pattern.

Part (b): Type of Model (E/P/I)

Essentially Correct (E)

Since ln(y) = a + bx is linear, this means y = e^(a+bx), which is an exponential model: y = e^1.2 × e^(0.045x).

Part (c): Prediction (E/P/I)

Essentially Correct (E)

ln(y-hat) = 1.2 + 0.045(30) = 1.2 + 1.35 = 2.55. y-hat = e^2.55 ≈ 12.81 grams.

Score Yourself:

Part (a):

Part (b):

Part (c):

How to Use This Page

Comparing Distributions

Part (a): Compare Distributions (E/P/I)

Part (b): Mean vs. Median (E/P/I)

Score Yourself:

Normal Distributions and z-Scores

Part (a): Proportion above 650 (E/P/I)

Part (b): 90th Percentile (E/P/I)

Part (c): Comparing z-scores (E/P/I)

Score Yourself:

Interpreting a Regression

Part (a): Slope (E/P/I)

Part (b): r² (E/P/I)

Part (c): Residual (E/P/I)

Part (d): Extrapolation (E/P/I)

Score Yourself:

Residual Plots and Model Assessment

Part (a): Residual Plot (E/P/I)

Part (b): High r² (E/P/I)

Score Yourself:

Designing an Experiment

Part (a): CRD (E/P/I)

Part (b): Why Random Assignment (E/P/I)

Part (c): Matched Pairs (E/P/I)

Score Yourself:

Sampling Bias and Study Design

Part (a): Identify Bias (E/P/I)

Part (b): Better Method (E/P/I)

Score Yourself:

Probability and Random Variables

Part (a): Expected Net Profit (E/P/I)

Part (b): Standard Deviation (E/P/I)

Part (c): 50 Games (E/P/I)

Score Yourself:

Binomial Probability

Part (a): Binomial Conditions (E/P/I)

Part (b): P(X = 18) (E/P/I)

Part (c): Doubt the Claim? (E/P/I)

Score Yourself:

Sampling Distributions

Part (a): Describe Sampling Distribution (E/P/I)

Part (b): Probability (E/P/I)

Part (c): Smaller Sample (E/P/I)

Score Yourself:

Sampling Distribution of a Proportion

Part (a): Conditions (E/P/I)

Part (b): Mean and SD (E/P/I)

Part (c): P(p-hat > 0.40) (E/P/I)

Score Yourself:

Significance Test for a Proportion

Part (a): Full 4-Step Procedure

Part (b): CI Connection (E/P/I)

Score Yourself:

Confidence Interval for a Proportion

Part (a): Construct CI (E/P/I)

Part (b): Interpret (E/P/I)

Part (c): Majority? (E/P/I)

Score Yourself:

One-Sample t-Test for a Mean

Part (a): Full 4-Step Procedure

Part (b): Type of Error (E/P/I)

Score Yourself:

Two-Sample t-Test

Part (a): Two-Sample t-test (E/P/I per step)

Part (b): CI Interpretation (E/P/I)

Score Yourself:

Chi-Square Test for Independence

Part (a): Hypotheses (E/P/I)

Part (b): Expected Count (E/P/I)

Part (c): Conclusion (E/P/I)

Part (d): Follow-up (E/P/I)

Score Yourself:

Chi-Square Goodness of Fit

Part (a): Hypotheses (E/P/I)

Part (b): Expected Counts (E/P/I)

Part (c): Chi-Square Statistic (E/P/I)

Score Yourself:

Regression Inference: t-Test for Slope

Part (a): Equation (E/P/I)