Lesson 3.3: Hypothesis Testing

Software Lab 3.3 Solutions

  1. H0: p = 0.62 (62%) versus HA: p ≠ 0.62.
  2. Conditions:
    • Independence: Yes, individuals likely responded independently in this random sample.
    • Random: Yes, a random sample was taken.
    • Success/failure condition: Yes, np0 = 60(0.62) = 37.2 ≥ 10 and nq0 = 60(0.38) = 22.8 ≥ 10.
    • 10% condition: Yes, n = 60 is less than 10% of US adults.
  3. Answers will vary, but if there are 39 “Yeses” and 21 “Noes”, then \hat{p} = 39/60 = 0.65 and Z = \dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0q_0}{n}}} = \dfrac{0.65-0.62}{\sqrt{\dfrac{(0.62)(0.38)}{60}}} = 0.479.
  4. Answers will vary, but if Z = 0.479, then the R code 2 * (1 - pnorm(0.479, mean=0, sd=1)) returns the p-value 0.6319.
  5. Answers will vary, but if the p-value is 0.6319, then 0.6319 > \alpha = 0.05 and we do not reject the null hypothesis.
  6. Answers will vary, but if we do not reject the null hypothesis, then there is insufficient evidence in the sample that the proportion of adults in the population that think climate change affects their local community is different from 62%.
  7. On average, we should get a p-value less than \alpha = 0.05 in 5% of repeated samples.
  8. A sample proportion less than 0.497 or greater than 0.743 (i.e., in one of the shaded tail areas) would result in a test statistic with a p-value less than \alpha = 0.05. These tail areas correspond to a probability of 0.05, which matches the answer to the previous question. The reason is because these tail areas represent the probability of a type 1 error, which is equal to the significance level \alpha.
  9. P(type 2 error) = 0.1341 and power = 0.8659.
  10. Increasing the sample size leaves P(type 1 error) unchanged, decreases P(type 2 error), and increases power.

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