Chapter 12: Inference for Proportions
Chapter 12: Significance Tests in Practice
|3. |The heights (in inches) of males in the United States are believed to be normally distributed with mean μ. The average height of|
| |a random sample of 25 American adult males is found to be [pic]= 69.72 inches, and the standard deviation of the 25 heights is |
| |found to be s = 4.15 inches. The standard error of [pic] is |
| |A) 0.17. B) 0.41. C) 0.69. D) 0.83. E) 2.04. |
|4. |Scores on the Math SAT (SAT-M) are believed to be normally distributed with mean μ. The scores of a random sample of three |
| |students who recently took the exam are 550, 620, and 480. A 95% confidence interval for μ based on these data is |
|A) |550.00 ± 173.88. |D) |550.00 ± 105.01. |
|B) |550.00 ± 142.00. |E) |550.00 ± 79.21. |
|C) |550.00 ± 128.58. | | |
Use the following to answer questions 5 through 8:
An SRS of 100 postal employees found that the average amount of time these employees had worked for the U.S. Postal Service was [pic] = 7 years, with a standard deviation of s = 2 years. Assume the distribution of the time the population of all postal employees has worked for the Postal Service is approximately normal with mean μ. Do the observed data represent evidence that μ has changed from its value of 7.5 years of 20 years ago? To determine this, we test the hypotheses H0: μ = 7.5, Ha: μ ≠ 7.5 using the one-sample t test.
|5. |The appropriate degrees of freedom for this test are |
| |A) 9. B) 10. C) 19. D) 99. E) 100. |
|6. |The P-value for the one-sample t test is |
|A) |larger than 0.10. |
|B) |between 0.05 and 0.10. |
|C) |between 0.01 and 0.05. |
|D) |below 0.01. |
|E) |impossible to determine, since the standard deviation of the study conducted 20 years ago is not given. |
|7. |A 95% confidence interval for the mean number of years μ that a current Postal Service employee has spent with the Postal |
| |Service is |
| |A) 7 ± 2. B) 7 ± 1.984. C) 7 ± 0.4. D) 7 ± 0.3. E) 7 ± 0.2. |
|8. |Suppose the mean and standard deviation we obtained were based on a sample of 25 postal workers, rather than 100. The P-value |
| |would be |
|A) |larger. |
|B) |smaller. |
|C) |unchanged, since the difference between [pic] and the hypothesized value μ = 7.5 is unchanged. |
|D) |unchanged, since both groups of workers have the same type of job. |
|E) |unchanged, since the variability measured by the standard deviation stays the same. |
|9. |We wish to see if the dial indicating the oven temperature for a certain model oven is properly calibrated. Four ovens of this |
| |model are selected at random. The dial on each oven is set to 300° F. After one hour, the actual temperature of each oven is |
| |measured. The temperatures observed are 305°, 310°, 300°, and 305°. Assuming that the actual temperatures for this model when |
| |the dial is set to 300° are normally distributed with mean μ, we test whether the dial is properly calibrated by testing the |
| |hypotheses |
| |H0: μ = 300, Ha: μ ≠ 300. Based on the data, the value of the one-sample t statistic is |
| |A) 5. B) 4.90. C) 2.82. D) 2.45. E) 1.23. |
Use the following to answer questions 10 and 11:
Bags of a certain brand of tortilla chips are claimed to have a net weight of 14 ounces. Net weights actually vary slightly from bag to bag and are normally distributed with mean μ. A representative of a consumer advocate group wishes to see if there is any evidence that the mean net weight is less than advertised and so intends to test the hypotheses H0: μ = 14, Ha: μ < 14.
To do this, he selects 16 bags of this brand at random and determines the net weight of each. He finds the sample mean to be [pic] = 13.88 ounces and the sample standard deviation to be s = 0.24 ounces.
|10. |Based on the data above, |
|A) |we would reject H0 at significance level 0.10 but not at level 0.05. |
|B) |we would reject H0 at significance level 0.05 but not at level 0.025. |
|C) |we would reject H0 at significance level 0.025 but not at level 0.01. |
|D) |we would reject H0 at significance level 0.01 but not at level 0.001. |
|E) |we would reject H0 at significance level 0.001. |
Use the following to answer questions 18 through 19:
The college newspaper of a large Midwestern university periodically conducts a survey of students on campus to determine the attitude on campus concerning issues of interest. Pictures of the students interviewed, along with quotes of their responses, are printed in the paper. Students are interviewed by a reporter “roaming” the campus who selects students to interview “haphazardly.” On a particular day the reporter interviews five students and asks them if they feel there is adequate student parking on campus. Four of the students say no. The sample proportion [pic] that respond “no” is thus 0.8.
|18. |Referring to the information above, the standard error of [pic] is |
| |A) 0.8. B) 0.64. C) 0.4. D) 0.18. E) 0.032. |
|19. |Referring to the information above, which of the following assumptions for inference about a proportion using a confidence |
| |interval are violated in this example? |
|A) |n is so large that both [pic] and n(1 -[pic]) are at least 10. |
|B) |The population is at least 10 times as large as the sample. |
|C) |We are interested in inference about a proportion. |
|D) |The data are an SRS from the population of interest. |
|E) |There appear to be no violations. |
Use the following to answer questions 20 and 21:
A radio talk show host with a large audience is interested in the proportion [pic]of adults in his listening area that think the drinking age should be lowered to 18. To find out, he poses the following question to his listeners: “Do you think that the drinking age should be reduced to 18 in light of the fact that 18-year-olds are eligible for military service?” He asks listeners to phone in and vote “yes” if they agree the drinking age should be lowered and “no” if not.
|20. |You are told that the sample proportion [pic] of those who phoned in and answered yes is [pic]= 0.70 and the standard error of|
| |the sample proportion is 0.0459. The number of people who phoned in |
|A) |is 21. |
|B) |is 50. |
|C) |is 100. |
|D) |is 200. |
|E) |cannot be determined from the information given. |
|21. |Of the 100 people who phoned in, 70 answered “yes.” Which of the following assumptions for inference about a proportion using a |
| |confidence interval are violated? |
|A) |The desired confidence level is not given. |
|B) |The population is at least 10 times as large as the sample. |
|C) |n is so large that both the count of successes [pic] and the count of failures n(1 – [pic]) are 10 or more. |
|D) |There appear to be no violations. |
|E) |The data are an SRS from the population of interest. |
|22. |Eighty rats whose mothers were exposed to high levels of tobacco smoke during pregnancy were put through a simple maze. The maze|
| |required the rats to make a choice between going left or going right at the outset. Sixty of the rats went right when running |
| |the maze for the first time. Assume that the 80 rats can be considered an SRS from the population all rats born to mothers |
| |exposed to high levels of tobacco smoke during pregnancy. (Note that this assumption may or may not be reasonable, but |
| |researchers often assume lab rats are representative of such larger populations since lab rats are often bred to have very |
| |uniform characteristics.) The standard error for the sample proportion [pic] of rats who went right the first time when running |
| |the maze is |
| |A) 0.0023. B) 0.0484. C) 0.0548. D) 0.0559. E) 0.4337. |
Use the following to answer questions 23 and 24:
A newspaper conducted a statewide survey concerning the 1998 race for state senator. The newspaper took a random sample (assume it is an SRS) of 1200 registered voters and found that 620 would vote for the Republican candidate. Let p represent the proportion of registered voters in the state that would vote for the Republican candidate.
|23. |Referring to the information above, a 90% confidence interval for p is |
| |A) 0.517 ± 0.014. B) 0.517 ± 0.022. C) 0.517 ± 0.024. D) 0.517 ± 0.028. |
| |E) 0.517 ± 0.249. |
|24. |Referring to the information above, what sample size would you need in order to estimate p with margin of error 0.01 with 95% |
| |confidence? Use the guess p = 0.5 as the value for p. |
| |A) 49. B) 1500. C) 4800. D) 4900. E) 9604. |
Use the following to answer questions 25 through 27:
A noted psychic was tested for ESP. The psychic was presented with 200 cards face down and asked to determine if the card featured one of five symbols: star, cross, circle, square, or three wavy lines. The psychic was correct in 50 cases. Let p represent the probability that the psychic correctly identifies the symbol on the card in a random trial.
|25. |Referring to the information above, and assuming that the 200 trials can be treated as an SRS from the population of all guesses|
| |the psychic would make in his lifetime a 95% confidence interval for p is |
|A) |0.25 ± 0.069. |
|B) |0.25 ± 0.060. |
|C) |0.25 ± 0.055. |
|D) |0.25 ± 0.050. |
|E) |We can assert that p = 0.20 with 100% confidence because the psychic is just guessing. |
|26. |Referring to the information above, suppose you wished to see if there is evidence that the psychic is doing better than if he |
| |were just guessing. To do this, you test the hypotheses H0: p = 0.20, Ha: p > 0.20. The P-value of your test is |
|A) |greater than 0.10. |D) |between 0.001 and 0.01. |
|B) |between 0.05 and 0.10. |E) |below 0.001. |
|C) |between 0.01 and 0.05. | | |
|27. |Referring to the information above, what sample size would you need in order to estimate p with margin of error 0.01 with 95% |
| |confidence? Use the guess p = 0.20 as the value for p. |
| |A) 1176. B) 4330. C) 6147. D) 7203. E) 9604. |
Answer Key
|1. |D |
|2. |A |
|3. |D |
|4. |A |
|5. |D |
|6. |C |
|7. |C |
|8. |A |
|9. |D |
|10. |B |
|11. |C |
|12. |A |
|13. |B |
|14. |E |
|15. |B |
|16. |D |
|17. |A |
|18. |D |
|19. |A |
|20. |D |
|21. |A |
|22. |D |
|23. |C |
|24. |C |
|25. |A |
|26. |D |
|27. |B |
|28. |C |
|29. |A |
|30. |C |
|31. |E |
|32. |D |
|33. |D |
|34. |E |
|35. |D |
|36. |B |
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