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Mixed skills
10 original practice questions · Untimed · Fractions and exact decimals accepted
1. A fitted line predicts y-hat = 11 + 2x. What response is predicted at x = 4?
Answer: __________________
aps-12-1-12. A dataset has mean 31. Each value is increased by 3. What is the new mean?
aps-03-1-13. A service claims its population satisfaction proportion equals 2/5. Which null hypothesis addresses that population claim?
aps-19-4-24. 24 volunteers are randomly assigned to a new tutoring method or the usual method. What does random assignment primarily support?
aps-05-1-15. Of 60 surveyed bus passengers, 20 choose option A. What fraction choose A? Give an exact fraction or decimal.
aps-02-4-26. A website invites clinic patients who feel strongly about a service to submit a rating. It receives 400 replies. Which concern remains?
aps-01-4-47. A study randomly samples from the target population and randomly assigns treatments. Which scope of conclusion is best supported?
aps-05-4-18. A one-sample t interval uses 15 independent normally distributed observations and estimates the unknown population SD. What are its degrees of freedom?
aps-21-2-39. Assume a normal model with mean 41 and SD 4. Using the 68–95–99.7 rule, approximately what percent lies between 33 and 49?
aps-03-4-110. A trial gives method A only to morning classes and method B only to evening classes, with 32 students per class. What is the main problem?
aps-05-2-3Answers & reasoning
1. 19
Substitute x = 4 into the fitted line. Compute 11 + 2 × 4. The predicted response is 19.
aps-12-1-12. 34 · Choice D
Adding a constant increases every observation equally. The mean shifts by the same constant. New mean = 31 + 3 = 34.
aps-03-1-13. H0: p = 2/5 · Choice D
Hypotheses describe a population parameter; the sample is evidence used to assess it. Here p denotes the fixed population satisfaction proportion. Why an alternative fails: State a hypothesis about the observed sample statistic rather than the unknown population parameter.
aps-19-4-24. A causal comparison of the methods in this experimental setting · Choice D
Treatments, rather than population members, are assigned randomly. Assignment helps distribute confounders between groups over repetitions. This supports a causal comparison within the study setting.
aps-05-1-15. 1/3 · Choice A
The relevant total is 60. Divide the count choosing A by this total: 20/60. The relative frequency is 1/3.
aps-02-4-26. Voluntary response can bias the result · Choice B
Participation is chosen by the respondents. People with strong opinions may be more likely to respond. Increasing the count does not remove voluntary response bias.
aps-01-4-47. Both population generalization and a causal comparison · Choice C
Assess how participants entered the study. Assess whether treatments were randomly assigned. Use sampling for population generalization and assignment for causal comparison.
aps-05-4-18. 14 · Choice B
Estimating the sample mean imposes one constraint on deviations from it. The n deviations sum to zero, leaving n−1=14 independent pieces. Why an alternative fails: Use n, ignoring the constraint introduced by estimating the sample mean.
aps-21-2-39. 95 · Choice C
Both endpoints are two standard deviations from the mean. The empirical rule assigns approximately 95% to this central interval. Enter 95, because the question asks for a percent.
aps-03-4-110. Method and time of day are confounded · Choice D
Method and class time change together. A difference may reflect either factor or both. The design cannot isolate method from time of day.
aps-05-2-3Original parameterized practice, independently checked by computation. Educator review pending. Not an official exam paper.