One-Proportion Hypothesis Testing Practice Problems ANSWERS
Perform an hypothesis test for each of the following. Be sure to (1) state your null and alternative hypotheses; (2) write your conclusion, which must include your decision (in terms of the null hypothesis), alpha level, p-value, and context (in terms of the alternative hypothesis). You may assume that conditions have been met and checked.
- A drug manufacturer claims that less than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300 out of 5000 Alzheimer’s patients whose families have given informed consent for the patients to participate in the study. In all, 25 of the subjects experience nausea. Use these data to perform a test of the drug manufacturer’s claim at the alpha = 0.05 significance level.
Ho: p = 0.10
Ha: p < 0.10
where p is the true, unknown population parameter/proportion of all patients like these who will experience nausea.
Fail to reject Ho. With a p-value of about 16% and an alpha level of 5%, we do not have sufficient evidence to say that the true proportion of all patients like these who will experience nausea is less than 10%.
- We hear that newborn babies are more likely to be boys than girls. Is this true? A random sample of 25,468 firstborn children included 13,173 boys. Do these data give convincing evidence that firstborn children are more likely to be boys than girls?
Ho: p = 0.50
Ha: p > 0.50
where p is the true, unknown population parameter/proportion of all first-born babies born that are boys.
Reject Ho. With a p-value of approximately 0 and an alpha level of 0.05, we have sufficient evidence to show that the true, unknown population parameter/proportion of all first-born babies born is greater than 50%.
- A state’s Division of Motor Vehicles claims that 60% of teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. Is there convincing evidence at the alpha = 0.05 significance level that the DMV’s claim is incorrect?
Ho: p = 0.60
Ha: p 0.60
where p is the true, unknown population parameter/proportion of all teens that pass their driving test on the first attempt.
Reject Ho. With a p-value of about 0.04 and an alpha level of 5%, we have sufficient evidence to determine that the proportion of all teens that pass their driving test on the first attempt is not equal to 60%.
NOW, WHAT IF.... OUR ALPHA LEVEL HAD BEEN 1% AND NOT 5%?? OUR DECISION WOULD HAVE CHANGED! OUR DECISION WOULD HAVE BEEN...
Fail to reject Ho. With a p-value of about 0.04 and an alpha level of 1%, we do not have sufficient evidence to determine that the proportion of all teens that pass their driving test on the first attempt is 0.60.
In either case, notice we DO NOT KNOW if the proportion of all teens that pass their driving test on the first attempt is greater than OR less than 60%; all we know is that the proportion is different than 60%! What would give us more information? What procedure?
- In a recent year, 73% of first-year college students responding to a national survey identified “being very well-off financially” as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important. Is there convincing evidence at the alpha level of 5% that the proportion of all first-year students at this university differs from the national value of 73%?
Ho: p = 0.73
Ha: p 0.73
where p is the true, unknown population parameter/proportion of all first-year students at this university who would consider ‘being very well-off financially” as an important personal goal.
Reject Ho. With a p-value of about 2.58% and an alpha level of 5%, we have sufficient evidence to determine that the proportion of all first-year students at this university who would consider ‘being very well-off financially’ as an important personal goal is not equal to 73%.
NOW, WHAT IF.... OUR ALPHA LEVEL HAD BEEN 1% AND NOT 5%?? OUR DECISION WOULD HAVE CHANGED! OUR DECISION WOULD HAVE BEEN...
Fail to reject Ho. With a p-value of about 0.025 and an alpha level of 1%, we do not have sufficient evidence to determine that the proportion of allfirst-year students at this university who would consider ‘being very well-off financially’ as an important personal goal is not equal to 73%.
Again, in either case, notice we DO NOT KNOW if the proportion of all first-year students at this university who would consider ‘being very well-off financially’ as an important personal goal is greater than OR less than 73%; all we know is that the proportion is different than 73%! What would give us more information? What procedure?
- Go to COC Math 140 Survey Data Spring 2016. Copy and paste “Gender Data” into StatCrunch. Let’s consider/assume that this is a SRS of community college students. We suspect that more females attend community colleges in the U.S. than males. Is there convincing evidence at the alpha level of 1% that the proportion of females attending community colleges in the U.S. is higher than the proportion of males who attend U.S. community colleges? (You can run an hypothesis test from raw data in a column like this – vs. using statistics. In StatCrunch, go to Stat, Proportion Stats, One Sample, With Data, select the data set (Gender Data), type ‘Female” in success, Ho p = 0.5 and Ha p > 0.5, compute.
Ho: p = 0.50
Ha: p > 0.50
where p is the true, unknown population parameter/proportion of all females who attend community colleges in the U. S..
Reject Ho. With a p-value nearly zero, and an alpha level of 1%, we have sufficient evidence to conclude that the true, unknown population parameter/proportion of all females who attend community colleges in the U.S. is greater than 50%.
NOW, WHAT IF.... OUR ALPHA LEVEL HAD BEEN 1% AND NOT 5%?? WOULD OUR DECISION CHANGE??
- Some experts claim that 11% of the world’s population has brown hair. We, however, strongly believe that this percentage is less than 11%. Go to COC Math 140 Survey Data 2016. Copy and paste “Hair Color Data” into StatCrunch. Let’s consider/assume that this is a SRS. Is there convincing evidence at the alpha level of 5% that the proportion of people in the world that have brown hair is less than 11%?
Ho: p = 0.11
Ha: p < 0.11
where p is the true, unknown population parameter/proportion of all people who have brown hair.
Fail to reject Ho. With a p-value of essentially1 and an alpha level of 5%, we do not have sufficient evidence to conclude that the true population proportion of all people who have brown hair is less than 11%.
- CHALLENGE: In early 2012, the Pew Internet and American Life Project asked a random sample of U.S. adults, “Do you ever... use Twitter or another service to share updates about yourself or to see updates about others?” According to Pew, the resulting 95% confidence interval is (0.123, 0.177). Does this interval provide convincing evidence, at an alpha level of 5%, that the actual proportion of U.S. adults who would say they use Twitter differs from 0.16? Justify your answer.