Lesson 3.2: Confidence Intervals

Software Lab 3.2 Solutions

  1. Answers will vary, but on average should be close to 62%.
  2. The sample proportion from another sample will generally not be identical, but should be somewhat similar since these are simple random samples of a reasonable size from the same population.
  3. Answers will vary, but if \hat{p}=0.667, say, then \hat{p} \pm z^* \times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}} = 0.667 \pm 1.960 \times \sqrt{\dfrac{0.667(1-0.667)}{60}} = 0.667 \pm 0.119 = (0.548, 0.786) = (54.8\%, 78.6\%).
  4. Answers will vary, but likely yes. The interval for \hat{p}=0.667 was calculated to be (54.8\%, 78.6\%), so this interval does contain the true population proportion, 62\%.
  5. 95% of those intervals would be expected to contain the true population proportion since this property follows from how confidence intervals are constructed.
  6. Answers will vary, but likely close to 0.95 or 95%. The proportion could differ from 0.95 since the samples are drawn randomly and the proportion only matches the confidence level “on average” (i.e., for an “infinite” number of samples).
  7. A 90% confidence interval is narrower than a 95% confidence interval (all else equal) because we can narrow the range of values if we’re less confident the interval contains the population proportion.
  8. Answers will vary, but if \hat{p}=0.667, say, then \hat{p} \pm z^* \times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}} = 0.667 \pm 1.645 \times \sqrt{\dfrac{0.667(1-0.667)}{60}} = 0.667 \pm 0.100 = (0.567, 0.767) = (56.7\%, 76.7\%). This 90% interval is narrower than the 95% interval from question 3.
  9. Answers will vary, but likely close to 0.90 or 90%, the confidence level of the intervals.
  10. A 99% confidence interval is wider than a 95% confidence interval (all else equal). If \hat{p}=0.667, say, then \hat{p} \pm z^* \times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}} = 0.667 \pm 2.576 \times \sqrt{\dfrac{0.667(1-0.667)}{60}} = 0.667 \pm 0.157 = (0.510, 0.824) = (51.0\%, 82.4\%).

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