Intro statistics · Final exam review
Statistics final exam review: practice test with answers
Your stats final probably covers the whole course, from describing data to regression. This review gives you a topic checklist, the six ideas that tie the course together, a formula sheet, four exam-style worked examples, and a 23-question mixed practice test that checks your answers instantly and shows full solutions.
Topic checklist
Most intro stats finals cover these six areas. The third column shows which practice test questions check each one, so after the test you know exactly what to go back to.
| Topic | You should be able to | Test questions | Review |
|---|---|---|---|
| Describing data and the normal model | Read shape, center and spread; spot potential outliers with 1.5 × IQR; turn values into z-scores and find normal areas | 1, 2, 3, 4 | Recap · Z score practice problems |
| Probability | Use the addition and multiplication rules, read a two-way table, test independence, and use the binomial and expected value | 5, 6, 7, 8, 9 | Recap |
| Sampling distributions and the CLT | Say what the central limit theorem promises and compute σ/√n | 10, 11, 12 | Recap |
| Confidence intervals and tests | Build and interpret a confidence interval, run a test, read a p-value and name Type I and Type II errors | 13, 14, 15, 16, 17, 20 | Recap · Hypothesis testing practice problems |
| Picking the right procedure | Match a scenario to the right procedure: one-sample, paired or two-sample t, proportion z, chi-square, ANOVA or regression | 18, 19 | Recap · Which statistical test should I use? |
| Regression and study design | Find and use a regression line, read r and r², compute a residual, and tell association from causation | 21, 22, 23 | Recap |
Swipe the table sideways to see every column.
The course in six ideas
Statistics courses feel like a pile of formulas, but the same few ideas run through all of them. If you understand these, most final exam questions come down to picking the right one.
1. Describe the data before you do anything else
Look at shape (symmetric or skewed), center and spread. In skewed data the mean gets pulled toward the long tail and the median doesn’t, so report the median and interquartile range (IQR) for skewed data and the mean and standard deviation for roughly symmetric data. Q1 and Q3 mark the lower and upper quarters of the data; IQR = Q3 − Q1. A value is a potential outlier by the usual rule if it is more than 1.5 × IQR below Q1 or above Q3.
A z-score says how many standard deviations a value is from the mean: z = (x − μ) / σ. For a normal model, z turns into an area with the table. The 68-95-99.7 rule is a quick check, not a replacement for the table.
2. Probability is counting with rules
“Or” adds (subtract the overlap). “And” multiplies, but only for independent events; otherwise use P(A and B) = P(A) × P(B | A). In a two-way table, “given” means you shrink to one row or column and use its total. Independent is not the same as mutually exclusive: mutually exclusive events can’t happen together, and if both have positive probability they can’t be independent.
The binomial model fits a fixed number of independent yes/no trials with the same success chance p. The expected value of a random variable is its long-run average, Σ x·P(x).
3. Statistics vary from sample to sample, and we can predict how
This is the bridge to everything after the midterm. A sample mean x̄ has its own distribution: centered at μ, with standard deviation σ/√n (the standard error). The central limit theorem says that for a large enough random sample, x̄ is approximately normal even if the population isn’t. A sample proportion p̂ works the same way with standard error √(p(1 − p)/n). Bigger samples mean less variation, but you need 4 times the sample to halve the error.
4. Intervals and tests use the same pieces
The common z and t confidence intervals here use estimate ± critical value × standard error. A z or t test uses (estimate − null value) / standard error. Other tests, such as chi-square, use a different statistic. Learn which shape fits before using the formula sheet.
The p-value is the chance of results at least as extreme as yours if H₀ were true. Reject H₀ when p ≤ α. Rejecting a true H₀ is a Type I error (its chance is α). Failing to reject a false H₀ is a Type II error. For a fixed sample size, lowering α makes Type II errors more likely. “Fail to reject” is not the same as “prove H₀”.
5. Picking the procedure is half the exam
Ask two questions: what kind of variable is the response (a number or a category), and how many groups, and are they paired or independent? Then use the table below or the full which statistical test chart.
| The data | Procedure |
|---|---|
| One mean, σ unknown | One-sample t interval or test |
| Same subjects measured twice, or matched pairs | Paired t on the differences |
| Means of two independent groups | Two-sample t |
| Means of 3 or more independent groups | One-way ANOVA (compare 3 or more means) |
| One proportion, or two proportions | One- or two-proportion z |
| Counts in categories (one or two categorical variables) | Chi-square goodness of fit or test of independence |
| Two quantitative variables | Correlation and linear regression |
6. Regression describes, design decides what you can claim
The least-squares line is ŷ = a + bx, with slope b = r(sy/sx) and intercept a = ȳ − b·x̄. The slope is the predicted change in y for each 1-unit increase in x. A residual is actual − predicted. r measures the strength and direction of a linear pattern, from −1 to 1, and r² is the share of the variation in y explained by the line. Don’t predict far outside the range of your x data (extrapolation).
What you can conclude depends on how the data were collected. Random sampling lets you generalize to the population. Random assignment to treatments (an experiment) lets you claim cause and effect. An observational study, however large, shows association only.
Formula sheet
Check what your instructor allows on exam day. Even if you get a formula sheet, the hard part is knowing which line to use, so the last column says when each one applies.
| What | Formula | Use it for |
|---|---|---|
| z-score | z = (x − μ) / σ | How unusual one value is |
| Outlier fences | Q1 − 1.5 × IQR and Q3 + 1.5 × IQR | Spotting potential outliers |
| Addition rule | P(A or B) = P(A) + P(B) − P(A and B) | “Or” questions |
| Conditional probability | P(A | B) = P(A and B) / P(B) | “Given” questions, two-way tables |
| Binomial | P(X = k) = C(n, k) pᵏ (1 − p)ⁿ⁻ᵏ; mean np; SD √(np(1 − p)) | Fixed n of independent yes/no trials |
| Expected value | E(X) = Σ x · P(x) | Long-run average of a random variable |
| Standard error of x̄ | σ / √n (use s / √n in practice) | Means |
| Standard error of p̂ | √(p̂(1 − p̂)/n) for intervals; √(p₀(1 − p₀)/n) for tests | Proportions |
| Confidence interval | estimate ± (critical value) × SE | Common z and t intervals |
| Test statistic | (estimate − null value) / SE | Every z or t test |
| Chi-square | χ² = Σ (O − E)² / E; df = (r − 1)(c − 1) for a two-way table | Counts in categories |
| Regression line | ŷ = a + bx, b = r(sy/sx), a = ȳ − b·x̄ | Predicting y from x |
| Residual | y − ŷ | How far a point is from the line |
Worked examples, final exam style
Final exam questions often chain two or three topics together. Each example below does that. Try each part before reading the solution. The data are made up.
Example 1: One battery vs. the average of many (normal model and CLT)
Phone battery life is roughly normal with mean 40 hours and standard deviation 6 hours.
(a) What share of single batteries last more than 46 hours?
z = (46 − 40) / 6 = 1. Area to the right of z = 1 is 1 − 0.8413 = 0.1587, about 16%.
(b) A store tests a random sample of 36 batteries. What is the chance their mean life is more than 41 hours?
Now the question is about x̄, so use the standard error: SE = 6 / √36 = 1 hour. z = (41 − 40) / 1 = 1. The chance is again 0.1587.
The point: 41 hours is only 1/6 of a standard deviation above the mean for one battery, but a full standard error above it for the average of 36. Averages vary much less than single values. Always ask “one value or a mean?” before you pick σ or σ/√n.
Example 2: A two-way table (conditional probability and independence)
A survey of 500 students records whether each one works a part-time job and whether they passed intro stats.
| Group | Passed | Failed | Total |
|---|---|---|---|
| Works | 150 | 50 | 200 |
| Doesn’t work | 255 | 45 | 300 |
| Total | 405 | 95 | 500 |
(a) P(passed) = 405 / 500 = 0.81.
(b) P(passed | works): restrict to the Works row. 150 / 200 = 0.75.
(c) Are working and passing independent? If they were, P(passed | works) would equal P(passed). It is 0.75 vs. 0.81, so in this sample they are not independent: students who work passed less often.
(d) P(works | failed): restrict to the Failed column. 50 / 95 ≈ 0.526. Notice this is not the same as P(failed | works) = 50 / 200 = 0.25. The order of “given” matters.
Example 3: Same data, interval and test (one proportion)
In a random sample of 400 students at one school, 232 say they study mostly the night before exams. Is that different from half?
(a) 95% confidence interval. p̂ = 232 / 400 = 0.58. SE = √(0.58 × 0.42 / 400) ≈ 0.0247. Margin = 1.96 × 0.0247 ≈ 0.048. Interval: 0.58 ± 0.048, or 0.532 to 0.628.
Interpretation: we are 95% confident that between about 53% and 63% of all students at this school study mostly the night before.
(b) Test H₀: p = 0.5 vs. Hₐ: p ≠ 0.5 at α = 0.05. Conditions: np₀ = n(1 − p₀) = 200, both at least 10. For the test, use p₀ (the null proportion) in the SE: √(0.5 × 0.5 / 400) = 0.025. z = (0.58 − 0.50) / 0.025 = 3.2. Two-sided p-value = 2 × P(Z > 3.2) ≈ 0.0014.
Conclusion. p ≤ 0.05, so reject H₀. There is strong evidence that more than half study mostly the night before. This agrees with the interval, which leaves out 0.5.
Watch out: the interval uses p̂ in the standard error, but the test uses p₀, because a test assumes H₀ is true.
Example 4: Regression from summary statistics
In an observational dataset of 50 students, hours studied (x) has mean 10 and SD 4. Final exam score (y) has mean 72 and SD 12. The correlation is r = 0.6.
(a) The regression line. b = r(sy/sx) = 0.6 × 12 / 4 = 1.8. a = ȳ − b·x̄ = 72 − 1.8 × 10 = 54. So ŷ = 54 + 1.8x.
(b) Interpret the slope. Each extra hour of studying goes with a predicted score 1.8 points higher, on average.
(c) Predict and find a residual. A student studied 15 hours and scored 75. ŷ = 54 + 1.8 × 15 = 81. Residual = 75 − 81 = −6. The student scored 6 points below the prediction.
(d) How much does the line explain? r² = 0.36, so 36% of the variation in scores is explained by the linear relationship with study hours.
(e) Can you say studying more causes higher scores? Not from these data. Students chose how much to study, so it is observational. Motivation or prior knowledge could drive both.
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Practice test: 23 mixed questions
This is a mixed test, like a real final: topics are not grouped and the questions don’t tell you which method to use. Assume random samples and that the usual conditions hold unless a question says otherwise. Type or choose an answer, press Check answer for instant feedback, and retry as often as you like. When you are done, press Score my test for a score by topic. Every question has a full solution.
Type numbers as decimals, like 0.25 or −3. A minus sign or hyphen both work.
Topic: Describing data and the normal model
Show solution 1
The mean is larger than the median. The few very long commutes pull the mean up, but the median only cares about the middle value. So in right-skewed data the mean is usually larger than the median.
Why the other choices are wrong:
- The mean is smaller than the median: That is the usual pattern for a tail to the left. A long right tail pulls the mean up.
- The mean and median are about equal: That is what you expect for a roughly symmetric shape, not a long tail.
- The median is pulled toward the tail: The median is resistant. It is the mean that gets pulled toward the long tail.
2. Exam times have Q1 = 20 minutes and Q3 = 34 minutes. Using the 1.5 × IQR rule, what is the upper fence for outliers (in minutes)?
Topic: Describing data and the normal model
Show solution 2
IQR = Q3 − Q1 = 34 − 20 = 14.
1.5 × IQR = 21.
Upper fence = Q3 + 21 = 55 minutes. Any time above 55 is a potential outlier by this rule and deserves a closer look. (The lower fence is 20 − 21 = −1, so no time can be a low outlier.)
3. Exam scores have mean 70 and standard deviation 8. What is the z-score of a student who scored 82?
Topic: Describing data and the normal model
Show solution 3
z = (x − mean) / SD = (82 − 70) / 8 = 12 / 8 = 1.5.
So 82 is 1.5 standard deviations above the mean.
4. Heights of adult women in a population are roughly normal with mean 64 inches and standard deviation 2.5 inches. What proportion are taller than 69 inches? Give a decimal to 4 places.
Topic: Describing data and the normal model
Show solution 4
z = (69 − 64) / 2.5 = 5 / 2.5 = 2.
From the table, the area to the left of z = 2 is 0.9772.
Area to the right = 1 − 0.9772 = 0.0228, so about 2.3% are taller than 69 inches.
5. For two events, P(A) = 0.4, P(B) = 0.3 and P(A and B) = 0.12. What is P(A or B)?
Topic: Probability
Show solution 5
Addition rule: P(A or B) = P(A) + P(B) − P(A and B).
= 0.4 + 0.3 − 0.12 = 0.58.
The overlap is subtracted once because it was counted in both P(A) and P(B).
Topic: Probability
Show solution 6
Yes, because P(A and B) = P(A) × P(B). P(A) × P(B) = 0.4 × 0.3 = 0.12, which equals P(A and B). So A and B are independent. Equivalently, P(A | B) = 0.12 / 0.3 = 0.4 = P(A).
Why the other choices are wrong:
- No, because A and B can happen at the same time: That is the test for mutually exclusive, not independent. Events that overlap can still be independent.
- No, because P(A) and P(B) are different: Independent events can have any probabilities. What matters is whether P(A and B) equals P(A) × P(B).
- You can’t tell without a two-way table: You already have enough. Compare P(A and B) with P(A) × P(B).
7. In a class of 120 students, 30 drink coffee daily and passed the final, 20 drink coffee daily and failed, 45 don’t drink coffee and passed, and 25 don’t drink coffee and failed. A student who passed is chosen at random. What is P(drinks coffee | passed)?
Topic: Probability
Show solution 7
| Group | Passed | Failed | Total |
|---|---|---|---|
| Coffee daily | 30 | 20 | 50 |
| No coffee | 45 | 25 | 70 |
| Total | 75 | 45 | 120 |
Given “passed”, the total is the Passed column: 75 students.
P(coffee | passed) = 30 / 75 = 0.4.
8. A multiple-choice quiz has 10 questions with 5 choices each. A student guesses independently on every question. What is the probability of exactly 2 correct? Give 3 decimal places.
Topic: Probability
Show solution 8
Each question is a success (right) with p = 1/5 = 0.2, independently, and n = 10 is fixed. So X ~ Binomial(10, 0.2).
P(X = 2) = C(10, 2) × 0.2² × 0.8⁸ = 45 × 0.04 × 0.16777 = 0.302.
9. A campus raffle ticket wins $10 with probability 0.1, wins $2 with probability 0.3, and loses $3 with probability 0.6. What is the expected value of one ticket, in dollars?
Topic: Probability
Show solution 9
E(X) = Σ x · P(x)
= 10(0.1) + 2(0.3) + (−3)(0.6) = 1 + 0.6 − 1.8 = −0.20.
Over many tickets, you lose about 20 cents per ticket.
10. IQ scores have mean 100 and standard deviation 15. For random samples of 25 people, what is the standard deviation of the sample mean (the standard error)?
Topic: Sampling distributions and the CLT
Show solution 10
SE(x̄) = σ / √n = 15 / √25 = 15 / 5 = 3.
Sample means of 25 people vary 5 times less than single scores.
Topic: Sampling distributions and the CLT
Show solution 11
Approximately normal, mean 40, standard deviation 2.5. By the central limit theorem, x̄ is approximately normal for a large enough sample, even from a skewed population. Its mean is μ = 40 and its standard deviation is σ / √n = 20 / 8 = 2.5.
Why the other choices are wrong:
- Skewed right like the population, standard deviation 20: With n = 64, the central limit theorem makes the sample means approximately normal, and their spread is σ/√n, not σ.
- Approximately normal, mean 40, standard deviation 20: The shape and center are right, but means vary less than single values: 20 / √64 = 2.5.
- Exactly normal, because n is more than 30: The CLT gives an approximation, not an exact normal shape. And the SD would still need to be found: 2.5.
Topic: Sampling distributions and the CLT
Show solution 12
4. SE = σ / √n. To halve SE, √n has to double, so n has to be 4 times as large.
Why the other choices are wrong:
- 2: SE = σ/√n. Doubling n only divides SE by √2 ≈ 1.41.
- 1/2: A smaller sample makes the standard error bigger, not smaller.
- √2: That would shrink SE by only about 16%. You need √n to double.
13. A random sample of 16 students sleeps a mean of 52 hours per week with standard deviation s = 8 hours. For a 95% t interval, t* = 2.131 with 15 degrees of freedom. What is the margin of error (in hours)? Give 2 decimal places.
Topic: Confidence intervals and tests
Show solution 13
SE = s / √n = 8 / √16 = 8 / 4 = 2.
Margin of error = t* × SE = 2.131 × 2 = 4.26 hours.
95% CI: 52 ± 4.26, or about 47.74 to 56.26 hours per week.
Topic: Confidence intervals and tests
Show solution 14
We are 95% confident the mean sleep of all students at the college is between 6.2 and 7.0 hours. A confidence interval estimates a population parameter, here the mean sleep of all students. “95% confident” means that the method captures the true mean in about 95% of samples.
Why the other choices are wrong:
- 95% of students at the college sleep between 6.2 and 7.0 hours: The interval is about the population mean, not about individual students. Many students sleep outside it.
- There is a 95% chance the sample mean is between 6.2 and 7.0: The sample mean is always in its own interval (it is the center). Confidence is about the population mean.
- 95% of all possible sample means fall between 6.2 and 7.0: A different sample would give a different interval. The 95% describes how often the method captures the true mean.
Topic: Confidence intervals and tests
Show solution 15
If H₀ were true, results at least this far from 100 would happen about 3% of the time. The p-value is the probability, assuming H₀ is true, of getting a test statistic at least as extreme as the one observed. A small p-value means the data would be unusual if H₀ were true.
Why the other choices are wrong:
- There is a 3% chance that H₀ is true: A p-value is computed assuming H₀ is true. It is not the probability that H₀ is true.
- There is a 97% chance that the alternative hypothesis is true: The p-value does not give the probability of either hypothesis.
- The true mean differs from 100 by 3%: The p-value says nothing about how big the difference is.
Topic: Confidence intervals and tests
Show solution 16
A Type I error becomes less likely and a Type II error becomes more likely. α is the chance of a Type I error, so lowering it makes Type I errors rarer. But it gets harder to reject H₀, so you fail to reject false null hypotheses more often. Type II error goes up and power goes down.
Why the other choices are wrong:
- Both errors become less likely: With the same n, there is a trade-off. Making it harder to reject H₀ means missing more real effects.
- A Type II error becomes less likely: Rejecting H₀ gets harder, so you miss real effects more often. Type II error goes up.
- The power of the test goes up: Power = 1 − P(Type II). Type II error goes up, so power goes down.
17. A college claims 30% of students use the library weekly. In a random sample of 200 students, 72 do. Test H₀: p = 0.30 against Hₐ: p ≠ 0.30. What is the z test statistic? Give 2 decimal places.
Topic: Confidence intervals and tests
Show solution 17
p̂ = 72 / 200 = 0.36.
Check: np₀ = 60 and n(1 − p₀) = 140, both at least 10.
SE = √(p₀(1 − p₀) / n) = √(0.30 × 0.70 / 200) = √0.00105 ≈ 0.0324.
z = (0.36 − 0.30) / 0.0324 ≈ 1.85.
Two-sided p-value ≈ 2 × 0.032 = 0.064. At α = 0.05 that is not enough evidence that the share differs from 30%.
Topic: Picking the right procedure
Show solution 18
Paired t test on the differences. Each student gives a before score and an after score, so the data come in pairs. Take after − before for each student and run a one-sample t test on those 30 differences.
Why the other choices are wrong:
- Two-sample t test: The two sets of scores come from the same students, so they are not independent groups. Analyze the differences.
- Chi-square test of independence: Scores are numbers, not counts in categories.
- One-way ANOVA: ANOVA compares independent groups, and there is really one sample of before-after pairs.
Topic: Picking the right procedure
Show solution 19
One-way ANOVA. One quantitative response (commute time) and one categorical factor with more than 2 groups (campus): that is one-way ANOVA.
Why the other choices are wrong:
- Two-sample t test: That compares only 2 groups. Running many t tests raises the overall chance of a Type I error.
- Chi-square test of independence: Commute time is a number. Chi-square works with counts in categories.
- Linear regression: Campus is a categorical variable with 4 groups, not a quantitative explanatory variable.
20. A chi-square test of independence uses a two-way table with 3 rows and 4 columns. How many degrees of freedom does the test have?
Topic: Confidence intervals and tests
Show solution 20
df = (r − 1)(c − 1) = (3 − 1)(4 − 1) = 2 × 3 = 6.
21. The least-squares line for predicting quiz score (out of 50) from hours studied is ŷ = 12 + 3.5x. A student who studied 8 hours scored 37. What is the residual?
Topic: Regression and study design
Show solution 21
Predicted: ŷ = 12 + 3.5 × 8 = 12 + 28 = 40.
Residual = y − ŷ = 37 − 40 = −3.
A negative residual means the point is below the line.
22. A scatterplot shows a negative linear relationship, and r² = 0.49. What is the correlation r?
Topic: Regression and study design
Show solution 22
|r| = √0.49 = 0.7.
The relationship is negative, so r = −0.7.
r² = 0.49 also means 49% of the variation in y is explained by the linear relationship with x.
Topic: Regression and study design
Show solution 23
Coffee and sleep are associated among adults, but this study can’t show that coffee causes less sleep. This is an observational study with a random sample. Random sampling supports generalizing the association to all adults. With no random assignment, lurking variables (stress, job hours) could explain the link, so no cause-and-effect claim.
Why the other choices are wrong:
- Coffee causes less sleep, because the sample was random: Random sampling lets you generalize to the population. Cause and effect needs random assignment to treatments.
- Nothing can be said about anyone outside the 1,000 people surveyed: The random sample does let you generalize the association to the population it was drawn from.
- It would be causal if the sample were larger: Sample size doesn’t fix lurking variables like stress or work hours. Only an experiment can.
Answer key
Open all 23 answers
- The mean is larger than the median. The few very long commutes pull the mean up, but the median only cares about the middle value. So in right-skewed data the mean is usually larger than the median.
- 55. IQR = 34 − 20 = 14. Upper fence = Q3 + 1.5 × IQR = 34 + 21 = 55 minutes.
- 1.5. z = (82 − 70) / 8 = 12 / 8 = 1.5. The score is 1.5 standard deviations above the mean.
- 0.0228. z = (69 − 64) / 2.5 = 2. The area to the right of z = 2 is 1 − 0.9772 = 0.0228.
- 0.58. P(A or B) = P(A) + P(B) − P(A and B) = 0.4 + 0.3 − 0.12 = 0.58.
- Yes, because P(A and B) = P(A) × P(B). P(A) × P(B) = 0.4 × 0.3 = 0.12, which equals P(A and B). So A and B are independent. Equivalently, P(A | B) = 0.12 / 0.3 = 0.4 = P(A).
- 0.4. There are 30 + 45 = 75 students who passed. Of those, 30 drink coffee. P(coffee | passed) = 30 / 75 = 0.4.
- 0.302. X is binomial with n = 10 and p = 0.2. P(X = 2) = C(10, 2)(0.2)²(0.8)⁸ = 45 × 0.04 × 0.1678 ≈ 0.302.
- −0.2. E(X) = 10(0.1) + 2(0.3) + (−3)(0.6) = 1 + 0.6 − 1.8 = −0.20. On average you lose 20 cents per ticket.
- 3. SE = σ / √n = 15 / √25 = 15 / 5 = 3.
- Approximately normal, mean 40, standard deviation 2.5. By the central limit theorem, x̄ is approximately normal for a large enough sample, even from a skewed population. Its mean is μ = 40 and its standard deviation is σ / √n = 20 / 8 = 2.5.
- 4. SE = σ / √n. To halve SE, √n has to double, so n has to be 4 times as large.
- 4.26. SE = s / √n = 8 / 4 = 2. Margin of error = t* × SE = 2.131 × 2 = 4.26 hours, so the interval is 52 ± 4.26, about 47.74 to 56.26 hours.
- We are 95% confident the mean sleep of all students at the college is between 6.2 and 7.0 hours. A confidence interval estimates a population parameter, here the mean sleep of all students. “95% confident” means that the method captures the true mean in about 95% of samples.
- If H₀ were true, results at least this far from 100 would happen about 3% of the time. The p-value is the probability, assuming H₀ is true, of getting a test statistic at least as extreme as the one observed. A small p-value means the data would be unusual if H₀ were true.
- A Type I error becomes less likely and a Type II error becomes more likely. α is the chance of a Type I error, so lowering it makes Type I errors rarer. But it gets harder to reject H₀, so you fail to reject false null hypotheses more often. Type II error goes up and power goes down.
- 1.85. p̂ = 72 / 200 = 0.36. SE = √(0.30 × 0.70 / 200) = 0.0324. z = (0.36 − 0.30) / 0.0324 ≈ 1.85.
- Paired t test on the differences. Each student gives a before score and an after score, so the data come in pairs. Take after − before for each student and run a one-sample t test on those 30 differences.
- One-way ANOVA. One quantitative response (commute time) and one categorical factor with more than 2 groups (campus): that is one-way ANOVA.
- 6. df = (rows − 1)(columns − 1) = (3 − 1)(4 − 1) = 2 × 3 = 6.
- −3. Predicted ŷ = 12 + 3.5(8) = 40. Residual = actual − predicted = 37 − 40 = −3. The student scored 3 points below the line.
- −0.7. r = −√0.49 = −0.7. The sign comes from the direction of the relationship, which is negative.
- Coffee and sleep are associated among adults, but this study can’t show that coffee causes less sleep. This is an observational study with a random sample. Random sampling supports generalizing the association to all adults. With no random assignment, lurking variables (stress, job hours) could explain the link, so no cause-and-effect claim.
A five-day review plan
- Day 1: Take the practice test cold. No notes. Press Score my test at the end and write down the topics where you missed questions.
- Day 2: Fix your weakest topic. Reread the recap above for it, redo the missed questions without looking, then work through the matching guide.
- Day 3: Fix the next topic, and build your own formula sheet. Write each formula with one line on when to use it.
- Day 4: Practice picking procedures. For every problem in your old quizzes and homework, name the procedure before you compute anything.
- Day 5: Retake the test against the clock. Aim for about 2 minutes a question, then review only what you missed.
FAQ
What is usually on an intro statistics final exam?
Most intro stats finals are cumulative. Expect describing data (center, spread, z-scores), probability and the binomial model, sampling distributions and the central limit theorem, confidence intervals, hypothesis tests, and regression. Many finals also ask you to pick the right procedure for a scenario. Your syllabus and old exams are the best guide to what your instructor stresses.
How should I study for a statistics final in a few days?
Take a mixed practice test first, without notes, to find your weak topics. Spend most of your time on those topics with worked problems, not rereading. Make your own one-page formula sheet as you go, then take a second mixed test under time pressure the day before.
When do I use t instead of z?
For a mean, use t when you estimate the population standard deviation with the sample standard deviation s, which is almost always. Use z for a mean only when σ is known. For a single proportion, intervals and tests use z.
What is the difference between a confidence interval and a hypothesis test?
An interval estimates a parameter with a range of plausible values. A test asks whether the data give convincing evidence against one specific value. They agree: a two-sided test at α = 0.05 rejects H₀ exactly when the 95% interval leaves out the null value, for the same t procedure on a mean.
Does a p-value of 0.03 mean there is a 3% chance the null hypothesis is true?
No. A p-value is computed assuming the null hypothesis is true. It is the chance of getting results at least as extreme as yours if H₀ were true. It is not the probability that H₀ is true.
Do I need to memorize formulas for a statistics final?
It depends on your course. Many instructors give a formula sheet or allow one you write yourself. Either way, you still need to know which formula fits which situation, so practice choosing the procedure, not just plugging in numbers.
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Sources and fact checks
Methods and rules were checked against these references. The made-up datasets and answers were recomputed. Normal areas come from the standard normal table; t* = 2.131 is the 95% critical value for 15 degrees of freedom. All datasets are made up.
- OpenStax Introductory Statistics 2e, 2.3: quartiles, IQR and the 1.5 × IQR outlier rule
- OpenStax 2.6: skewness and the mean and median
- OpenStax 3.4: contingency tables and conditional probability
- OpenStax 4.3: the binomial distribution
- OpenStax 7.1: the central limit theorem for sample means
- OpenStax 8.2: t intervals for a mean
- OpenStax 9.6: tests for a single mean and a single proportion
- OpenStax 9.2: Type I and Type II errors
- OpenStax 11.3: chi-square tests of independence
- OpenStax 1.4: observational studies and experiments
- OpenStax 12.3: the regression equation
- Penn State STAT 200: Elementary Statistics