Confidence Intervals and Introduction to Inference Multiple Choice Practice Problems for AP Statistics (page 2)

By — McGraw-Hill Professional
Updated on Feb 4, 2011


  1. The correct answer is (e).
  2. .

    You need a sample of at least n = 323.

  3. The correct answer is (e).
  4. The correct answer is (a). It is not significant at the .01 level because .015 is greater than .01.
  5. The correct answer is (b). n = 16 df = 16 – 1 = 15. Using a table of t distribution critical values, look in the row for 15 degrees of freedom and the column with 0.05 at the top (or 90% at the bottom). On a TI-84 with the invT function, the solution is given by invT(0.95,15).
  6. The correct answer is (c). Because 0 is not in the interval (0.07, 0.19), it is unlikely to be the true difference between the proportions. III is just plain wrong! We cannot make a probability statement about an interval we have already constructed. All we can say is that the process used to generate this interval has a 0.95 chance of producing an interval that does contain the true population proportion.
  7. The correct answer is (d). I is not correct since you cannot make a probability statement about a found interval—the true mean is either in the interval (P = 1) or it isn't (P = 0). II is correct and is just a restatement of "Intervals constructed by this procedure will capture the true mean 99% of the time." III is true—it's our standard way of interpreting a confidence interval. IV is true. Since the interval constructed does not contain 0, it's unlikely that this interval came from a population whose true mean is 0. Since all the values are positive, the interval does provide statistical evidence (but not proof ) that the program is effective at promoting weight loss. It does not give evidence that the amount lost is of practical importance.
  8. The correct answer is (d). To reject the null at the 0.01 level of significance, we would need to have z < –2.33.
  9. The correct answer is (d). A confidence level is a statement about the procedure used to generate the interval, not about any one interval. It's difficult to use the word "probability" when interpreting a confidence interval and impossible, when describing an interval that has already been constructed. However, you could say, "The probability is 0.99 that an interval constructed in this manner will contain the true population proportion."
  10. The correct answer is (c). For df = 15 – 1 = 14, t* = 2.264 for a 96% confidence interval (from Table B; if you have a TI-84 with the invT function, invT(0.98,14)=2.264. The interval is 74.5 ± (2.264) = 74.5 ± 1.871.
  11. The correct answer is (e). Because we are concerned that the actual amount of coverage might be less than 400 sq ft, the only options for the alternative hypothesis are (d) and (e) (the alternative hypothesis in (a) is in the wrong direction and the alternatives in ((b) and (c) are two-sided). The null hypothesis given in (d) is not a form we would use for a null (the only choices are =, ≤, or ≥). We might see H0 : μ = 400 rather than H0 : μ ≥ 400. Both are correct statements of a null hypothesis against the alternative HA : μ < 400.
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