Free practice final
Statistics practice final (free, with answers)
A typical intro statistics final covers describing data, study design, probability, the normal and binomial distributions, sampling distributions, confidence intervals, hypothesis tests, and regression. Any table values you need are given in the question, so all you need is a calculator. Try each question on your own before opening the answer, then read the explanation to check your reasoning.
Topics
- Descriptive Statistics
- Study Design
- Probability
- Probability Distributions
- Sampling Distributions
- Confidence Intervals
- Hypothesis Tests
- Correlation and Regression
15 questions
Try each one before you open its answer.
1. Sale prices of homes in a town are mostly between $200,000 and $350,000, but a few mansions sold for over $3 million. Which pair of summary statistics best describes the typical price and its spread?
- A Mean and standard deviation
- B Median and interquartile range (IQR)
- C Mean and interquartile range (IQR)
- D Median and range
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Answer: B. Median and interquartile range (IQR)
The median and IQR are resistant: a few extreme prices barely move them. The mean, standard deviation, and range are all pulled toward the mansions, so they overstate a typical price and its spread.
2. In a randomized experiment comparing a new study app with a standard one, what is the main purpose of randomly assigning students to the two groups?
- A To make the results apply to all college students
- B To keep students from knowing which app they are using
- C To make sure both groups have exactly the same number of students
- D To balance other variables between the groups, so differences in outcome can be attributed to the app
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Answer: D. To balance other variables between the groups, so differences in outcome can be attributed to the app
Random assignment spreads lurking variables such as motivation or prior grades evenly across the groups on average, which is what allows a cause-and-effect conclusion. Generalizing to a population comes from random sampling, and hiding the treatment from subjects is blinding.
3. Find the sample standard deviation s of the data 4, 8, 6, 2, 10.
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Answer: s = √10 ≈ 3.16 (rounded to two decimals)
The mean is 6, and the squared deviations are 4, 4, 0, 16, and 16, which sum to 40. Dividing by n − 1 = 4 gives a sample variance of 10, so s = √10 ≈ 3.16.
4. Events A and B have P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.2. Which statement is true?
- A A and B are mutually exclusive.
- B A and B are independent, and P(A or B) = 0.9.
- C A and B are independent, and P(A or B) = 0.7.
- D A and B are not independent, and P(A or B) = 0.7.
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Answer: C. A and B are independent, and P(A or B) = 0.7.
They are independent because P(A) × P(B) = 0.4 × 0.5 = 0.2, which equals P(A and B). By the addition rule, P(A or B) = 0.4 + 0.5 − 0.2 = 0.7. They are not mutually exclusive, since P(A and B) is not 0.
5. Scores on a final exam are approximately normal with mean 70 and standard deviation 8. About what proportion of students score above 82? (Use P(Z < 1.0) = 0.8413, P(Z < 1.5) = 0.9332, and P(Z < 2.0) = 0.9772.)
- A 0.0668
- B 0.9332
- C 0.1587
- D 0.0228
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Answer: A. 0.0668
The z-score is (82 − 70) / 8 = 1.5. The area above it is 1 − 0.9332 = 0.0668, so about 6.7% of students score above 82.
6. A survey of 200 students finds that 80 are commuters and 120 live on campus. Of the commuters, 60 have a part-time job; of the students living on campus, 30 do. If a surveyed student with a part-time job is chosen at random, what is the probability that the student is a commuter?
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Answer: 2/3 ≈ 0.667 (rounded to three decimals)
Conditioning on having a job restricts attention to the 60 + 30 = 90 students with jobs. Of those, 60 are commuters, so P(commuter | job) = 60/90 = 2/3.
7. A student guesses randomly on all 10 questions of a multiple-choice quiz, and each question has 4 options. Let X be the number of correct answers. What are the mean and standard deviation of X?
- A Mean 2.5, standard deviation ≈ 1.88
- B Mean 2.5, standard deviation ≈ 0.43
- C Mean 7.5, standard deviation ≈ 1.37
- D Mean 2.5, standard deviation ≈ 1.37
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Answer: D. Mean 2.5, standard deviation ≈ 1.37
X is binomial with n = 10 and p = 0.25, so the mean is np = 2.5. The standard deviation is √(np(1 − p)) = √1.875 ≈ 1.37; 1.875 itself is the variance.
8. The amounts customers spend on a single visit to a store are strongly right-skewed, with mean $50 and standard deviation $30. For random samples of 100 visits, which best describes the sampling distribution of the sample mean x̄?
- A Strongly right-skewed like the population, with mean $50 and standard deviation $30
- B Approximately normal, with mean $50 and standard deviation $3
- C Approximately normal, with mean $50 and standard deviation $30
- D Approximately normal, with mean $50 and standard deviation $0.30
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Answer: B. Approximately normal, with mean $50 and standard deviation $3
By the central limit theorem, x̄ is approximately normal for a sample this large even though the population is skewed. Its mean is μ = $50 and its standard deviation is σ/√n = 30/√100 = $3.
9. A random sample of 25 students has a mean test score of x̄ = 82 with standard deviation s = 10. Using t* = 2.064 (df = 24), construct a 95% confidence interval for the population mean score.
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Answer: About 77.87 to 86.13 (82 ± 4.13, rounded to two decimals)
The margin of error is t* × s/√n = 2.064 × 10/√25 = 2.064 × 2 = 4.128. The interval is 82 − 4.128 to 82 + 4.128.
10. From a random sample, a 95% confidence interval for the mean commute time of a city's workers is 22 to 28 minutes. What does "95% confidence" mean here?
- A 95% of the city's workers have commutes between 22 and 28 minutes.
- B 95% of the commute times in the sample are between 22 and 28 minutes.
- C If many random samples were taken and an interval computed from each in the same way, about 95% of those intervals would contain the true mean commute time.
- D A new random sample would have a 95% chance of producing a sample mean between 22 and 28 minutes.
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Answer: C. If many random samples were taken and an interval computed from each in the same way, about 95% of those intervals would contain the true mean commute time.
The 95% describes the method: intervals built this way capture the true mean in about 95% of samples. The interval estimates the population mean, not where individual commute times fall.
11. A study tests H₀: μ = 100 against Hₐ: μ ≠ 100 and gets a p-value of 0.03. Which statement correctly interprets the p-value?
- A If μ really equals 100, there is about a 3% chance of getting a result at least as extreme as the one observed.
- B There is a 3% probability that H₀ is true.
- C There is a 97% probability that Hₐ is true.
- D The true mean differs from 100 by about 3%.
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Answer: A. If μ really equals 100, there is about a 3% chance of getting a result at least as extreme as the one observed.
A p-value is computed assuming H₀ is true: it is the probability of data at least as extreme as what was observed. It is not the probability that either hypothesis is true.
12. A coin is flipped 400 times and lands heads 220 times. To test H₀: p = 0.5 against Hₐ: p ≠ 0.5, compute the one-proportion z test statistic, using the standard error based on p₀.
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Answer: z = 2.00
The sample proportion is 220/400 = 0.55, and the standard error under H₀ is √(0.5 × 0.5 / 400) = 0.025. So z = (0.55 − 0.50) / 0.025 = 2.00, which gives a two-sided p-value of about 0.046.
13. A clinical trial tests H₀: a new drug does not lower blood pressure, against Hₐ: it does lower blood pressure. In this context, a Type I error would be:
- A Concluding the drug does not lower blood pressure when it actually does
- B Concluding the drug lowers blood pressure when it actually does
- C Not concluding the drug lowers blood pressure when it actually does not
- D Concluding the drug lowers blood pressure when it actually does not
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Answer: D. Concluding the drug lowers blood pressure when it actually does not
A Type I error is rejecting a true H₀, here declaring the drug effective when it is not. The first choice is a Type II error (failing to reject a false H₀), and the other two are correct decisions.
14. For a sample of used cars, the correlation between age (in years) and resale price is r = −0.8. Which statement is correct?
- A About 64% of the variation in resale price is explained by the linear relationship with age.
- B About 80% of the variation in resale price is explained by the linear relationship with age.
- C Resale price drops by $0.80 for each additional year of age.
- D The relationship is weak, because r is negative.
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Answer: A. About 64% of the variation in resale price is explained by the linear relationship with age.
The coefficient of determination is r² = (−0.8)² = 0.64, the share of variation in y explained by the regression on x. The sign of r gives the direction, not the strength, and r is not the slope.
15. A regression of final exam score (y) on hours studied (x) gives ŷ = 52 + 4.5x. A student who studied 6 hours scored 83. What is this student's residual?
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Answer: 4 points (observed 83 − predicted 79)
The predicted score is 52 + 4.5 × 6 = 79, and a residual is observed minus predicted: 83 − 79 = 4. A positive residual means the student scored higher than the line predicts.