Liberty Union High School District / Overview



MC Practice – Inference for Proportions/CI1. On the day before an election in a large city, each person in a random sample of 1,000 likely voters is asked which candidate he or she plans to vote for. Of the people in the sample, 55 percent say they will vote for candidate Taylor. A margin of error of 3 percentage points is calculated. Which of the following statements is appropriate?(A) The proportion of all likely voters who plan to vote for candidate Taylor must be the same as the proportion of voters in the sample who plan to vote for candidate Taylor (55 percent), because the data were collected from a random sample.(B) The sample proportion minus the margin of errors is greater than 0.50, which provides evidence that more than half of all likely voters plan to vote for candidate Taylor.(C) It is not possible to draw any conclusions about the proportions of all likely voters who plan to vote for candidate Taylor because the 1,000 likely voters in the sample represent only a small fraction of all likely voters in a large city.(D) It is not possible to draw any conclusions about the proportions of all likely voters who plan to vote for candidate Taylor because this is not an experiment.(E) It is not possible to draw any conclusions about the proportions of all likely voters who plan to vote for candidate Taylor because this is a random sample and not a census.2. A newspaper poll found that 52 percent of the respondents in a large random sample of likely voters in a district intend to vote for candidate Smith rather than the opponent. A 95 percent confidence interval for the population proportion was computed to be 0.520.04. Based on the confidence interval, which of the following should the newspaper report to its readers?(A) Smith will win because a majority of voters are in favor of Smith.(B) There is a 95% chance that Smith will win.(C) The poll predicts Smith will win, but there is a 5% chance that the prediction is incorrect due to sampling error.(D) With 95% confidence, there is convincing evidence that Smith will win.(E) No prediction about who will win can be made with 95% confidence.3. A school administrator is interested in estimating the proportion of students in the district who participate in community service activities. From a random sample of 100 students in the district, the administrator will construct a 99 percent confidence interval for the proportion of all district students who participate in community service activities. Which of the following statements must be true?(A) The population proportion will be in the confidence interval.(B) The probability that the confidence interval will include the population proportion is 0.99.(C) The probability that the confidence interval will include the sample proportion is 0.99.(D) The population proportion and the sample proportion will be equal. (E) The probability that the population proportion and the sample proportion will be equal is 0.99. 4. A polling agency conducted a survey by selecting 100 random samples, each consisting of 1,200 United States citizens. The citizens in each sample were asked whether they were optimistic about the economy. For each sample, the polling agency created a 95 percent confidence interval for the proportion of all United States citizens who were optimistic about the economy. Which of the following statements is the best interpretation of the 95 percent confidence interval?(A) With 100 confidence intervals, we can be 95% confident that the sample proportion of citizens of the United States who are optimistic about the economy is correct.(B) We would expect about 95 of the 100 confidence intervals to contain the proportion of all citizens of the United States who were optimistic about the economy.(C) We would expect about 5 of the 100 confidence intervals to not contain the proportion of all citizens of the United States who were optimistic about the economy.(D) Of the 100 confidence intervals, 95 of the intervals will be identical because they were constructed from samples of the same size of 1,200.(E) The probability is 0.95 that 100 confidence intervals will yield the same information about the sample proportion of citizens of the United States who were optimistic about the economy.5. Jessica wanted to determine if the proportion of males for a certain species of laboratory animal is less than 0.5. She was given access to appropriate records that contained information on 12,000 live births for the species. To construct a 95 percent confidence interval, she selected a simple random sample of 100 births from the records and found that 31 births were male. Based on the study, which of the following expressions is an approximate 95 percent confidence interval estimate for p, the proportion of males in the 12,000 live births?(A) 0.31±1.96(0.31)(0.69)12000 (C) 0.31±1.96(0.5)(0.5)12000(E) 0.31±1.96(0.31)(0.69)100(B) 0.31±1.645(0.31)(0.69)12000 (D) 0.31±1.645(0.5)(0.5)1006. A random sample of 300 students is selected from a large group of students who use a computer-equipped classroom on a regular basis. Occasionally, students leave their USB drive in a computer. Of the 300 students questioned, 180 said that they write their name on their USB drive. Which of the following is a 98 percent confidence interval for the proportion of all students using the classroom who write their name on their USB drive?(A) 0.4±2.33(0.4)(0.6)300(C) 0.6±2.33(0.4)(0.6)300(E) 0.6±2.05(0.4)(0.6)300(B) 0.4±1.96(0.4)(0.6)300(D) 0.6±1.96(0.4)(0.6)3007. The management team of a company with 10,000 employees is considering installing charging stations for electric cars in the company parking lots. In a random sample of 500 employees, 15 reported owning an electric car. Which of the following is a 99 percent confidence interval for the proportion of all employees at the company who own an electric car?(A) 0.03±2.326(0.03)(0.97)500 (B) 0.15±2.326(0.15)(0.85)500(C) 0.03±2.576(0.03)(0.97)500(D) 0.15±2.576(0.15)(0.85)500(E) 0.03±2.576(0.03)(0.97)500+(0.03)(0.97)10,0008. As part of a national sleep survey, a random sample of adults was selected and surveyed about their physical activity and the number of hours they sleep each night. Of the 183 adults who exercised regularly (exercisers), 59 percent reported sleeping at least seven hours at night. Of the 88 adults who did not exercise regularly (non-exercisers), 52 percent reported sleeping at least seven hours at night. Which of the following is the most appropriate standard error for a conference interval for the difference in proportions of adults who sleep at least seven hours at night among exercisers and non-exercisers?(A) (0.59)(0.41)183+(0.52)(0.48)88(B) 0.590.41+(0.52)(0.48)183+88 (C)(0.57)(0.43)(1183+188)(D)(0.5)(0.5)(1183+188)(E)(0.5)(0.5)(1+1183+88)9. A marketing company wants to estimate the proportion of consumers in a certain region of the country who would react favorably to a new marketing campaign. Further, the company wants the estimate to have a margin of error no more than 5 percent with a 90 percent confidence. Of the following, which is closest to the minimum number of consumers needed to obtain the estimate with the desired precision?(A) 136(B) 271(C) 385(D) 542(E) 76910. A 90 percent confidence interval is to be created to estimate the proportion of television viewers in a certain area who favor moving broadcast of the late weeknight news to an hour earlier than it is currently. Initially, the confidence interval will be created using a simple random sample of 9,000 viewers in the area. Assuming that the sample proportion does not change, what would be the relationship between the width of the original confidence interval and the width of a second 90 percent confidence interval that is based on a sample of only 1,000 viewers in the area?(A) The second confidence interval would be 9 times as wide as the original confidence interval.(B) The second confidence interval would be 3 times as wide as the original confidence interval.(C) The width of the second confidence interval would be equal to the width of the original confidence interval.(D) The second confidence interval would be 1/3 times as wide as the original confidence interval.(E) The second confidence interval would be 1/9 times as wide as the original confidence interval.11. A commercial for breakfast cereal is shown during a certain television program. The manufacturer of the cereal wants to estimate the percent of television viewers who watch the program. The manufacturer wants the estimate to have a margin of error of at most 0.02 at a level of 95 percent confidence. Of the following, which is the smallest sample size that will satisfy the manufacturer’s requirements?(A) 40(B) 50(C) 100(D) 1,700(E) 2,50012. The manager of a car company will select a random sample of its customers to create a 90 percent confidence interval to estimate the proportion of its customers who have children. Of the following, which is the smallest sample size that will result in a margin of error no more than 6 percentage points?(A) 100(B) 125(C) 150(D) 200(E) 275MC Practice – Inference for Proportions/HT13. A researcher is conducting a study of charitable donations by surveying a simple random sample of households in a certain city. The researcher wants to determine whether there is a convincing statistical evidence that more than 50 percent of households in the city gave a charitable donation in the past year. Let p represent the proportion of all households in the city that gave a charitable donation in the past year. Which of the following are appropriate hypotheses for the researcher?(A) Ho: p = 0.5 and Ha: p > 0.5(B) Ho: p = 0.5 and Ha: p ≠ 0.5(C) Ho: p = 0.5 and Ha: p < 0.5(D) Ho: p > 0.5 and Ha: p ≠ 0.5(E) Ho: p > 0.5 and Ha: p = 0.514. An environmental scientist wants to test the null hypothesis that an antipollution device for cars is not effective. Under which of the following conditions would a Type I error be committed?(A) The scientist concludes that the antipollution device is effective when it actually is not.(B) The scientist concludes that the antipollution device is not effective when it actually is.(C) The scientist concludes that the antipollution device is effective when it actually is.(D) The scientist concludes that the antipollution device is not effective when it actually is not.(E) A Type I error cannot be committed in this situation.15. A university will add fruit juice vending machines to its classroom buildings if the student body president is convinced that more than 20 percent of the students will use them. A random sample of n students will be selected and asked whether or not they would use the vending machines. A large-sample test for proportions at the significance level of α = 0.05 will be performed. The null hypothesis that the proportion of all students who would use the vending machines is 20 percent will be tested against the alternative that more than 20 percent of all students would use them. For which of the following situations would the power of the test be highest?(A) The sample size is n=750, and 20 percent of all students use the vending machines.(B) The sample size is n=750, and 25 percent of all students use the vending machines.(C) The sample size is n=1000, and 25 percent of all students use the vending machines.(D) The sample size is n=500, and 50 percent of all students use the vending machines.(E) The sample size is n=1000, and 50 percent of all students use the vending machines.16. A one-sided hypothesis test is to be performed with a significance level of 0.05. Suppose that the null hypothesis is false. If a significance level of 0.01 were to be used instead of a significance level of 0.05, which of the following would be true?(A) Neither the probability of a Type II error nor the power of the test would change.(B) Both the probability of a Type II error and the power of the test would decrease.(C) Both the probability of a Type II error and the power of the test would increase.(D) The probability of a Type II error would decrease and the power of the test would increase.(E) The probability of a Type II error would increase the power of the test would decrease.17. A medical doctor uses a diagnostic test to determine whether a patient has arthritis. A treatment will be prescribed only if the doctor thinks the patient has arthritis. The situation is similar to using a null and an alternative hypothesis to decide whether to prescribe the treatment.Ho: The patient does not have arthritis.Ha: The patient has arthritis.Which of the following represents a Type II error for the hypothesis?(A) Diagnosing arthritis in a patient who has arthritis.(B) Failing to diagnose arthritis in a patient who has arthritis.(C) Diagnosing arthritis in a patient who does not have arthritis.(D) Failing to diagnose arthritis in a patient who does not have arthritis.(E) Prescribing treatment to a patient regardless of the diagnosis.18. A representative of a car manufacturer in the United States made the following claim in a news report:“Ten years ago, only 53 percent of Americans owned American-made car, but that figure is significantly higher today.”A research group conducted a study to investigate whether the claim was true. The group found that 56 percent of a randomly selected sample of car owners in the United States owned American-made cars. A test of the appropriate hypotheses resulted in a p-value of 0.283. Assuming the conditions for inference are met, is there sufficient evidence to conclude, at the significance level of α=0.05, that the proportion of all car owners in the United States who own American-made cars has increased from what it was ten years ago?(A) Yes, because 0.56 > 0.53(B) Yes, because a reasonable interval for the proportion is 0.56 0.283(C) Yes, because 0.56 - 0.53 = 0.03 and 0.03 < 0.05(D) No, because 0.283 < 0.53(E) No, because 0.283 > 0.0519. In the states of Florida and Colorado, veterinarians investigating obesity in dogs obtained random sample of pet medical records and recorded the weight of the dogs in the sample. A test was concluded of Ho: p1 = p2 versus Ha: p1 ≠ p2, where p1 represents the proportion of all overweight dogs in Florida and p2 represents the proportion of all overweight dogs in Colorado. The resulting test statistics for a two-sample z-test for a difference between proportions was 1.85. At the significance level α=0.05, which of the following is a correct conclusion?(A) There is not sufficient statistical evidence to conclude that the proportion of all overweight dogs in Florida is different from the proportion of all overweight dogs in Colorado because the p-value is greater than 0.05.(B) There is not sufficient statistical evidence to conclude that the proportion of all overweight dogs in Florida is different from the proportion of all overweight dogs in Colorado because the z-test statistics is greater than 0.05.(C) There is sufficient statistical evidence to conclude that the proportion of all overweight dogs in Florida is different from the proportion of all overweight dogs in Colorado because the p-value is greater than 0.05.(D) There is sufficient statistical evidence to conclude that the proportion of all overweight dogs in Florida is different from the proportion of all overweight dogs in Colorado because the p-value is less than 0.05.(E) There is sufficient statistical evidence to conclude that the proportion of all overweight dogs in Florida is greater than the proportion of all overweight dogs in Colorado because the z-test statistics is positive.20. A pollster is interested in comparing the proportion of women and men in a particular town who are in favor of a ban on fireworks within town borders. The pollster plans to test the hypothesis that the proportion of women in favor of the ban is different from the proportion of men in favor of the ban. There are 4,673 women and 4,502 men who live in the town. From a simple random sample of 40 women in the town, the pollster finds that 38 favor the ban. From an independent simple random sample of 50 men in the town, the pollster finds that 27 favor the ban. Which of the following statements is true about this situation?(A) Because the samples are from normal population, a two-proportion z-test would be valid.(B) Because the size of each sample is greater than 30, a two-proportion z-test would be valid.(C) Because the number who favored the ban is greater than 10 in both groups, a two-proportion z-test would be valid.(D) Because of the relative sizes of the populations and samples, a two-proportion z-test would be valid.(E) A two-proportion z-test would not be valid for these data.21. Dan selected a random sample of 100 students from the 1,200 at his school to investigate preferences for making up school days lost due to emergency closings. The results are shown in the table below.Dan incorrectly performed a large sample test of the difference in two proportions using 58100 and 42100 and calculated a p-value of 0.02. Consequently, he concluded that there was a significant difference in preference for the two options. Which of the following best describes his error in the analysis of these data?(A) No statistical test was necessary because 0.58 is clearly larger than 0.42.(B) The results of the test were invalid because less than 10% of the population was sampled.(C) Dan performed a two-tailed test and should have performed a one-tailed test.(D) A one-sample test for a proportion should have been performed because only one sample was used.(E) More options should have been included, and a chi-square test should have been performed. ................
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