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Statistics and Data Analysis Checkpoint

A free Algebra I lesson from the “Statistics and Data Analysis” unit, with a worked example and practice problems including step-by-step solutions.

This checkpoint samples Scatter Plots and Trend Lines, Line of Best Fit, Correlation Coefficient, and Two-Way Tables. The 25 review questions and 25 scored questions were authored separately. Lines of fit are supplied, sometimes by two points on the line; you do not need to calculate a regression. A residual is observed minus predicted: positive means the point is above the line (underprediction), and negative means below (overprediction). Interpolation uses an x-value between the smallest and largest data x-values; extrapolation goes beyond them and needs extra caution. The correlation coefficient satisfies −1 ≤ r ≤ 1. Its sign gives direction and its magnitude gives strength of linear association; compare magnitudes without using fixed weak/strong cutoffs. A curved relationship can have r near zero. Observational data alone cannot establish causation; random assignment can support a causal comparison under suitable conditions. In a two-way table, joint and marginal relative frequencies use the grand total, while the words ‘of the …’ name the group for a conditional frequency. Compare conditional rates to assess association. Percent answers are exact unless rounding is explicitly requested: when asked to enter a percent, enter 65 or 65%, not 0.65. All scored data are included in each question; review practice also includes two scatter plots. Results describe this sample of proposed outcome A1.11, not an external certification. The three written tasks use rubrics, are not automatically scored, and do not change your quiz score.

What you'll learn

Why it matters: Data can guide decisions about delivery costs, resources, travel, and customer choices. A useful analysis states what was observed, which group a percentage describes, and where a prediction or causal claim goes beyond the evidence.

Worked example

Problem. A fitted delivery model is y = 1.4x + 6 minutes for journeys of 3 to 15 km. For a 10 km journey, the actual time was 23 minutes. Predict the time, find the residual, and interpret it.

  1. Use x = 10: predicted y = 1.4(10) + 6 = 20 minutes.
  2. Residual = observed − predicted = 23 − 20 = +3 minutes.
  3. The trip took 3 minutes longer than predicted, so its point is above the line and the model underpredicts this trip.
  4. 10 km is between 3 and 15 km, so this is interpolation. Even within the observed inputs, a fitted line need not predict every observation exactly.

Answer: Predicted 20 minutes; residual +3 minutes; the model underpredicts

Lesson preview: this page shows 5 practice items. The interactive activity contains 25 practice items and 25 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.

Practice problems

1. Measurements (x, y) are (1, 4), (2, 7), (3, 9), (4, 13), (5, 15), (6, 18). Which description best fits the pattern?

Choices: Positive and approximately linear, with points close to a line · Negative and approximately linear, with points close to a line · Positive and strongly curved, with a progressively steeper pattern · No clear association, with points scattered without a pattern

Show solution
  1. As x increases from 1 to 6, y increases from 4 to 18.
  2. The increases in y are similar but not identical, so the points lie close to an upward straight line.
  3. This is a positive, approximately linear association; the tight pattern indicates a strong linear association.

Answer: Positive and approximately linear, with points close to a line

2. A lab records (x, y): (0, 2), (1, 3), (2, 6), (3, 11), (4, 18). Which model shape is most appropriate, and why?

Choices: A curve: the y-increases become larger at equal x-steps · A straight line: every one-unit x-step adds the same y-amount · A horizontal line: the y-values stay close to one fixed value · No model: any set of five measurements has no usable pattern

Show solution
  1. At one-unit x-steps, the y-increases are 1, 3, 5, and 7.
  2. Those increasing changes make a curved pattern, rather than a roughly straight cluster.
  3. A positive association can be nonlinear; an upward pattern alone does not justify a line.

Answer: A curve: the y-increases become larger at equal x-steps

3. Machine age in years and resale value in hundreds of dollars are paired as (1, 31), (2, 28), (3, 26), (4, 22), (5, 20). Should a line summarize these observations?

Choices: Yes; similar decreases support an approximate linear model · No; a line is allowed only if every point lies exactly on it · Yes; five observations guarantee accurate long-term predictions · No; a line can describe only quantities that increase together

Show solution
  1. Successive changes in resale value are −3, −2, −4, and −2 hundreds of dollars.
  2. The decreases are reasonably similar, so a line can summarize the negative association.
  3. The observations need not lie exactly on the line, and the fit does not guarantee predictions far beyond the observed ages.

Answer: Yes; similar decreases support an approximate linear model

4. Most sensor readings cluster near a line. One reading sits far above that cluster and far to its right. What is the best next step before using a fitted line?

Choices: Check the reading and compare fits with and without it · Delete the reading because every outlier is a recording error · Keep the fit because one reading cannot change a fitted line · Report a stronger association because an outlier always helps

Show solution
  1. A point far from the others, especially in its x-value, can affect a fitted line substantially.
  2. Check whether the measurement is valid and investigate why it differs.
  3. Compare the fits rather than assuming that the association must become stronger or weaker; do not discard valid data without justification.

Answer: Check the reading and compare fits with and without it

5. Across towns, library funding and adult literacy are associated. Town income was not recorded. Which explanation should be investigated before claiming that library spending alone caused the difference?

Choices: Town income may affect both library funding and literacy · The association proves that library funding determines literacy · A literacy difference proves that income has no relevant effect · More towns would make this association alone establish causation

Show solution
  1. This is a comparison of existing towns, not a randomized experiment.
  2. Town income is a plausible lurking variable: it may relate to both public resources and educational opportunities.
  3. The association alone cannot isolate the causal effect of library spending.

Answer: Town income may affect both library funding and literacy

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Graph observations and assess a proposed fit

A delivery study records (packages, minutes): (2, 15), (4, 19), (6, 26), (8, 30), (10, 35). Plot the observations with labeled axes. Draw the supplied fit y = 2.5x + 10, interpret its rate and intercept, and predict for 7 packages. Explain why predicting for 40 packages requires caution and why the intercept may not describe a real zero-package trip.

Review rubric and solution
  • Plots all five observations with packages on the horizontal axis and minutes on the vertical axis.
  • Draws y = 2.5x + 10 using two or more correct points on the line.
  • Interprets 2.5 as predicted additional minutes per package and 10 as predicted minutes at zero packages.
  • Computes 27.5 minutes at 7 packages and identifies interpolation.
  • Explains that 40 packages and zero packages lie outside the measured inputs; neither prediction is validated by these data.
  1. Plot the five observations. Draw the fitted line using, for example, (2, 15) and (10, 35).
  2. The rate is 2.5 predicted minutes per additional package. The intercept predicts 10 minutes at zero packages, an input not observed here.
  3. For 7 packages, y = 2.5(7) + 10 = 27.5 minutes, an interpolation.
  4. For 40 packages, the equation gives 110 minutes, but this extrapolation may fail if delivery conditions change. The zero-package intercept likewise needs contextual evidence.

Use residuals to evaluate a model

At x = 1, 2, 3, 4, 5, observed y-values are 17, 19, 17, 23, 29. A supplied fit is y = 3x + 11. Make a table of predictions and residuals (observed − predicted). Plot residual against x, interpret the residual at x = 3, and discuss whether the sign pattern suggests a limitation of the straight-line fit.

Review rubric and solution
  • Computes predictions 14, 17, 20, 23, 26.
  • Computes residuals +3, +2, −3, 0, +3 with observed-minus-predicted order.
  • Plots residuals against x with a horizontal zero reference line and labels both axes.
  • Explains that at x = 3 the model overpredicts by 3 units.
  • Notes possible structure in the residuals without claiming that five observations prove a particular alternative model.
  1. Predictions are 14, 17, 20, 23, 26. Subtract these from observations to obtain 3, 2, −3, 0, 3.
  2. Plot (1, 3), (2, 2), (3, −3), (4, 0), (5, 3) on a residual plot with a labeled zero line.
  3. At x = 3, the observation is 3 units below the prediction, so the model overpredicts.
  4. Positive residuals toward both ends and a negative residual in the middle suggest possible curvature. With only five observations, investigate further rather than concluding that a specific curve is established.

Complete a table and critique a causal claim

A recycling survey has rows received a leaflet/no leaflet and columns recycle/do not recycle. Counts are 48, ? / 30, 50. There are 160 people, with 80 in each row. Complete the table with row and column totals. Compare recycling percentages within the rows. A report says the leaflet caused every difference. Critique that claim and suggest a stronger study.

Review rubric and solution
  • Finds the missing count 32 and column totals 78 and 82, with grand total 160.
  • Computes 48/80 = 60% and 30/80 = 37.5% within the respective rows.
  • Identifies an association with a difference of 22.5 percentage points.
  • Distinguishes association from causation because leaflet receipt was not stated to be randomly assigned.
  • Suggests random assignment of leaflet distribution with comparable conditions and follow-up measurement.
  1. The missing count is 80 − 48 = 32. The recycling column totals 48 + 30 = 78, and the other column totals 32 + 50 = 82; together they give 160.
  2. Recycling rates are 60% with a leaflet and 37.5% without, a difference of 22.5 percentage points.
  3. This is an association, but leaflet recipients may differ in other ways. The survey alone cannot establish that the leaflet caused the difference.
  4. Randomly assigning leaflets and then measuring recycling under comparable conditions would provide stronger evidence of a causal effect.

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