Chapter 10 Practice Problems

Transcription

1 Chapter 10 Practice Problems Name 1. The standard deviation of SAT scores is 100 points. A researcher decides to take a sample of 500 students' scores to estimate the mean score of students in your state. What is the standard deviation of the sample mean? A. 0. B C. 5 D. 100 E. Can't determine without the sample mean. The 99.7% confidence interval for the mean length of frog jumps is (1.64 cm, cm). Which of the following statements is a collect interpretation of 99.7% confidence? A. Of the total number of frogs in your area of the country, 99.7% can jump between 1.64 cm and cm. B. There is a 99.7% chance that the true mean length of frog jumps falls between 1.64 cm and cm. C. If we were to repeat this sampling many times, 99.7% of the confidence intervals we could construct would contain the true population mean. D. 99.7% of the confidence intervals we could construct after repeated sampling would go from 1.64 cm to cm E. There's a 99.7% chance that any particular frog I catch can jump between 1.64 cm and cm. 3. True or False: A 95% confidence interval is narrower than a 90% confidence interval for the same data set. 4. What's the critical z-value for an 85% confidence interval? A..803 B C D. Can't be determined without knowing the population standard deviation E. Can't be determined without knowing the sample size 5. What's z* for a 78% confidence interval? A..77 B C D. 1. E. Cannot compute without the standard deviation. 6. A researcher computes a 90% confidence interval for the mean weight (in lb) of widgets produced in a factory. The interval is (7., 8.9). Which of these is a correct interpretation of this interval? A. Out of all the widgets produced in all widget factories, 90% weigh between 7. and 8.9 lb. B. We can be 90% confident that all the widgets weigh between 7. and 8.9 lb C. There's a 90% chance the population value is between 7. and 8.9 lb. D. Ninety percent of all sample means are equivalent to the true mean weight of all the widgets. E. If you drew many samples of size n and constructed a confidence interval from each sample, 90% of the intervals would contain the true population value.

2 7. A teacher administers a standardized math test to his class of 75 students. The mean score (out of 300 possible points) is 35. From previous studies, you know the population standard deviation is 8. Using the sample data given, calculate a 95% confidence interval for the population mean, A. (34.1,35.9) B. (6.7, 43.3) C. (8.7,41.3) D. (33.0, 37.0) E. (00.0, 300.0) Questions 8-9 refer to the following information: A researcher is interested in estimating the mean blood alcohol content (BAC) of people arrested for driving under the influence. The sample consists of 50 individuals with a mean BAC of.145. Based on past data, the researcher assumes a population standard deviation of What's the margin of error for a 90% confidence interval in this scenario? A B C..107 D E. Not enough information to compute the margin of error 9. What's the 95% confidence interval for the scenario above? A. (.137,.153) B. (.080,.10) C. (.138,.15) D. (.111,.17) E. Not enough information to compute the interval 10. A teacher administers a standardized math test to his class of 75 students. The mean score (out of 300 possible points) is 35. From previous studies, you know the population standard deviation is 8. The principal has decided that she wants to estimate the average score to within 4 points (margin of error = 4) with 99% confidence. If she can only administer the test to one random sample of students, how large should this sample be to achieve the desired margin of error and confidence level? A. 75 students B. 16 students C. 188 students D. 36 students E. 35 students 11. A random sample of 85 adults found that average calorie consumption was,100 per day. Previous research has found a standard deviation of 450 calories, and you use this value for. Construct a 99% confidence interval for the population mean. A. (1,905.5,,89.6) B. (,004.4,,195.6) C. (,097.4,,10.6) D. (1,650.0,,550.0) E. (1,974.3,,5.7)

3 A random sample of 85 adults found that average calorie consumption was,100 per day. Previous research has found a standard deviation of 450 calories, and you use this value for. A researcher wants to estimate a 95% confidence interval and is willing to accept a margin of error of 50 calories. She knows it will cost $50 to survey each member of the sample. Given this information, how much will it cost to survey the minimum number of people? A. $4,050 B. $6,870 C. $15,600 D. $3,750 E. $15, Which of the following statements is true? A. Smaller sample sizes produce larger margins of error because smaller samples always have larger standard deviations. B. The point estimate is the measure of variation used in the computation of the margin of error. C. Smaller sample sizes produce larger margins of error because smaller samples are more susceptible to random variability. D. All of the above are correct. E. None of the above are correct. 14. Sulfur compounds cause off-odors in wine, so winemakers want to know the odor threshold, the lowest concentration of a compound that the human nose can detect. The odor threshold for dimethyl sulfide (DMS) in trained wine tasters is about 5 micrograms per liter of wine ( ). The untrained noses of consumers may be less sensitive, however. Here are the DMS odor thresholds for 10 untrained students: Assume that the standard deviation of the odor threshold for untrained noses is known to be = 7. A. Find z* for a 93% confidence interval for the true mean of the population. Show your work using the graph provided. B. Construct and interpret a 93% confidence interval for the mean DMS odor threshold among all untrained students. C. Nick explains that 93% confident means that the process you used to calculate the confidence interval would result in an interval that captures the sample mean in 93% of all possible samples of 10 untrained testers from the population. Comment on Nick s interpretation of the confidence level. D. Determine the sample size you would need to estimate µ within at a 93% confidence level About 4,000 high school students took the AP Statistics exam in 001. The freeresponse section of the exam consisted of five open-ended problems and an investigative task. Each free-response question is scored on a 0 to 4 scale (with 4 being the best). A random sample of 5 student papers yielded the following scores on one of the freeresponse questions:

4 A. Is a sample of 5 papers large enough to provide a good estimate of the mean score of all 4,000 students on this exam problem? Justify your answer. B. Do you think the population of scores on this question is Normally distributed? Explain why or why not. C. Construct and interpret a 95% confidence interval for the mean score on this exam question. Be sure to explain why it s okay to calculate the interval in light of your answer to Question. 16. A news article on a Gallup Poll noted that 8 percent of the 1548 adults questioned felt that those who were able to work should be taken off welfare. The article also said, The margin of error for a sample size of 1548 is plus or minus three percentage points. A. Opinion polls usually announce margins of error for 95% confidence. Using this fact, explain to someone who knows no statistics what margin of error plus or minus three percentage points means. B. Do you agree with the margin of error stated in the poll results? If so, explain why. If not, tell why not. C. What is the 99% confidence interval from this poll? Interpret the interval in context. D. What conditions must be met in order for the confidence interval in Question 3 to be valid? Check whether each of those conditions is satisfied in this case. E. This poll was conducted by telephone. Explain how undercoverage could lead to a biased estimate in this case.

5 Chapter 10 ANSWERS 1. Answer B To calculate the standard deviation of a sample mean, use the formula, n which is 100, or approximately 4.47 points Answer C. if we were to repeat this sampling many times, 99.7% of the confidence intervals we could construct would contain the true population mean. This means that 99.7% of the time the sample mean will come within 3 standard deviations of the population mean. Therefore, 99.7% of the time the confidence interval we construct will cover the true population mean, and 0.3% of the time we'll miss it. 3. Answer False. A 95% confidence interval is wider than a 90% confidence interval because we need to include more possible values in our estimate to increase our confidence that we've captured the population mean. 4. Answer B Using the standard normal curve, you want to find the two z-values between which 85% of the area lies. This means that 7.5% of the area falls below the lower z-value, and 7.5% falls above the upper z-value. 5. Answer D. 1.. In a 78% confidence interval, we're interested in the area that represents the middle 78% of the normal curve. You need to find the z-value that gives the distance on either side of the mean to create this area. You can find this z-value by finding the z-score for the lowertail probability C + (1- C)/, where C is your confidence level expressed as a decimal. 6. Answer E. If you drew many samples of size n and constructed a confidence interval from each sample, 90% of the intervals would contain the true population value. This means we're 90% confident the interval contains the true population value. A confidence interval is a range of values that should contain the true population value. 7. Answer C. (8.7,41.3). Using the formula x z* 95% confidence level z* = 1.96., = 8, x = 35, and n = 75. For a n 8. Answer B The margin of error = z* times the standard deviation of the sample mean. In this case, the z* for a 90% confidence interval is and the standard deviation of the.065 sample mean is = = Thus (1.645)(.0041) = n Answer A. (.137,.153). Construct the interval using (estimate) (margin of error). The estimate is the sample mean of.145. The margin of error for a 95% confidence level is z*( ) = (1.96)(.0041) =.008. This gives you the interval.137 to.153. x 10. Answer D. 36 students. Using the formula n = whole number to get 36. * z m.576(8) =, you get the next 4

6 Answer E. (1,974.3,,5.7). The margin of error in this case is z* =.576 = n Since the mean is,100 calories, the interval is to,5.7 calories. * z 1.96(450) 1. Answer C. $15,600. Using n = =,, you get Round up m 50 number to get 31. This gives you 31($50.00) = $15, Answer C. smaller sample sizes produce larger margins of error because smaller samples are more susceptible to random variability. Conversely, larger samples produce smaller margins of error because they are less susceptible to random variability a. z* = (The closest value.9649 in table A) b. P: Estimate unknown mean DMS odor threshold among all untrained students. A: SRS- doesn t say sample is random, hence proceed with caution since samply may not provide a good estimate of population mean Normality- check graph 7 T: = (6.39,34.41) 10 C: We are 93% confident that the mean DMS odor threshold for all untrained students is between 6.39 and c. Nick is incorrect. We are trying to capture the population mean. The interpretation of the confidence level is that we got these numbers by a method that captures the true parameter (in this case the population). d. z* m n n n n a. n is a little small there will be a lot of variability associated with the estimate. b. No, the histogram of the sample data is heavily skewed to the right. c. P: Want, mean score on this Exam Q for all AP Students in 001. A: unknown hence this is a cue to use sample t-interval.

7 SRS- Since a random sample was selected we can take these results as representative sample from this population Normality- Rightly skewed but not outliers hence it is ok! Independence- these 5 randomly selected scores should be independent measurements T: = (0.57, 1.51); degree of freedom (df) = 4 5 C: We are 95% confident that the actual mean score on this FRQ for all AP Students in 001 was between 0.57 and a. The margin of error shows how accurate we believe the estimate for the unknown parameter is, based on the variability of the estimate. In 95% of all possible samples of this size from this population, our estimate will be 3% of the true population percentage. b =.04 which does NOT equal c. pˆ margin of error = = (.560,.3093) 1548 We are 99% confident that the actual proportion of all adults who feel that those who are able to work should be taken off welfare is between.5 and.31 d. We would like to assume that the Gallup poll used a random sample, but we don t know if it was an SRS. The counts of successes (agreed) & failures (disagreed) are both at least 10 (433 and 1115) and the number of adults is greater than 10(1548) = Conditions are satisfied. e. People who do not have a phone would not be represented in the sample. These are usually individuals with lower incomes, some of whom may be on welfare. These people may have similar views on the issue being addressed in the survey, but the estimate obtained from the survey would tend to be biased in the opposite direction due to under coverage. This type of non-sampling error is not included in the margin of error.