Chapter 16 stats
A pharmaceutical company tests whether a drug lifts the headache relief rate from the 25% achieved by the placebo. They fail to reject the null hypothesis because the P-value is 0.465. Further testing shows that the drug actually relieves headaches in 38% of people. A. The company made a Type I error. The null hypothesis was not rejected, but it was false. The true relief rate was greater than 0.25. B. The company made a Type II error. The null hypothesis was not rejected, but it was false. The true relief rate was greater than 0.25. Your answer is correct. C. The company did not make an error. They did not reject the null hypothesis when the P-value was almost 0.5. D. The company made a Type II error. It used a P-value to determine its conclusion instead of a critical value. E. The company made a Type I error. It used a P-value to determine its conclusion instead of a critical value
B. The company made a Type II error. The null hypothesis was not rejected, but it was false. The true relief rate was greater than 0.25.
Type II error
Failing to reject the null hypothesis when its false / null hyp should have been rejected. ex: 0.8<0.7
Type I error
Rejecting null hypothesis when it is true/null hypothesis should not have been rejected. ex: 15>14.9
If the company's lawyers are worried about being sued for selling an unsafe product, should they make the value of α smaller or larger
They should make the value of α smaller. This means a smaller chance of declaring the stands safe when they are not actually safe.
Have harsher penalties and ad campaigns increased seat-belt use among drivers and passengers? Observations of commuter traffic have failed to find evidence of a significant change compared with three years ago. Explain what the study's P-value of 0.55 means in this context. A. The P-value 0.55 is the probability of getting results like the ones obtained in the study by natural variation alone, assuming that the proportion of commuters who wear seat belts has increased. B. The P-value 0.55 is the probability that the proportion of commuters who wear seat belts has not changed. C. The P-value 0.55 is the probability of getting results like the ones obtained in the study by natural variation alone, assuming that the proportion of commuters who wear seat belts has not changed. Your answer is correct. D. The P-value 0.55 is the probability that the proportion of commuters who wear seat belts has changed. E. One minus the P-value, or 0.45, is the probability that the proportion of commuters who wear seat belts has changed.
The P-value 0.55 is the probability of getting results like the ones obtained in the study by natural variation alone, assuming that the proportion of commuters who wear seat belts has not changed.
A test of H0: p=0.3 vs. HA: p<0.3 fails to reject the null hypothesis. Later it is discovered that p=0.4.
no error occurred
construct a 98% confidence interval for the sample proportion
statcrunch > proportion stats > # of successes, # observations, confidence interval
A test of H0: μ=15 vs. HA: μ>15 rejects the null hypothesis. Later it is discovered that μ=14.9.
type I error, should have not been rejected/fail to reject
A test of H0: p=0.8 vs. HA: p<0.8 fails to reject the null hypothesis. Later it is discovered that p=0.7.
type II error, should have been rejected
