ECO 578 Midterm Exam | Latest complete solution, Worth 300 points.

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ECO 578 Midterm Exam 1. For a binomial probability experiment, with n=150 and p=.1, we can use the normal approximation to the binomial distribution even without continuity correction. (Ch6) 2. ... The probability of an event is the sum of the probabilities of the sample space outcomes that correspond to the event. (Ch4) 3. We may need to perform the continuity correction even if the population is 20 times or more than the sample size. (Ch6) 4. The sampling distribution of the sample mean is always normally distributed according to the Central Limit Theorem (Chaps. 6 and 7) 5. If the population is normal and its standard deviation is known, then the t- distribution is appropriate for a sample size of 20. (Ch8) 6. For a continuous distribution, Probability of (X greater than or equal to 10) is greater than the probability of (X greater than 10) (Ch6) 7. If the population is normally distributed with known variance then the sample mean may not be normally distributed for a very small sample size. (Ch7) - 8. When the level of confidence and sample standard deviation remain the same, a confidence interval for a population mean based on a sample of n=100 will be narrower than a confidence interval for a population mean based on a sample of n=50. (Ch8) 9. If we examine selected items from the population, we are not conducting a census of the population. (Ch1) 10. When the level of confidence and sample proportion p remain the same, a confidence interval for a population proportion p based on a sample of n=100 will be wider than a confidence interval for p based on a sample of n=150. (Ch8) 11. Events whose union has the probability equal to 1 are called Exhaustive Events.(CH4) 12. The t distribution is symmetrical like the standard normal curve and has the same mean but its spread or variance is less than the standard normal distribution. (Ch8) 13. When constructing a confidence interval for a sample proportion, the t distribution is always appropriate if the sample size is small. (Ch8) 14. An estimator is called consistent if its variance and standard deviations consistently remain the same regardless of changes in the sample size.(Ch7) 15. If the population is normally distributed then the sample mean is also normally distributed even for small sample size. (Ch7) 16. The reason sample variance has a divisor of n-1 rather than n is that it makes the standard deviation an unbiased estimate of the population standard deviation. (Chs. 3 and 7) 17. When determining the sample size n, if the value found for n is 79.2, we would choose to sample 79 observations. (Ch8) Multiple Choices (7 points each) 1. In a study conducted by UCLA, it was found that 25% of college freshmen support increased military spending. If 6 college freshmen are randomly selected, find the probability that: Exactly 3 support increased military spending. (Ch5) A. 0.0330 B. 0.1318 C. 0.7844 D. 0.9624 2. In a statistic class, 10 scores were randomly selected with the following results were obtained: 74, 73, 77, 77, 71, 68, 65, 77, 67, and 66. What is the median? (Ch3) A. 71.5 B. 72 C. 77 D. 71 E. 73 3. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent more than 90 days overdue. The historical records of the company show that over the past 8 years 14 percent of the accounts are delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What is the probability that at least 30 accounts will be classified as delinquent? (Ch 6) A. 31.86% B. 18.14% C. 81.86% D. 63.72% E. 75.84% 4. According to a hospital administrator, historical records over the past 10 years have shown that 20% of the major surgery patients are dissatisfied with after-surgery care in the hospital. A scientific poll based on 400 hospital patients has just been conducted. What is the probability that at least 70 patients will not be satisfied with the after-surgery care? (Ch6) A. 89.44% B. 39.44% C. 10.56% D. 78.88% E. 84.49% 5. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. A random sample of 16 hospitals in one state had a mean LOS for non-heart patients in 2000 of 5 days and a standard deviation of 2 days. Calculate a 95% confidence interval for the population mean LOS for non-heart patients in the state's hospitals in 2000. (Ch8) A. [4.02 5.98] B. [3.93 6.07] C. [4.18 5.82] D. [3.52 6.47] E. [3.71 6.29] 6. A person's telephone area code is an example of a(n) _____________ variable. (Ch1) A. Nominal B. Ordinal C. Interval D. Ratio E. Continuous 7. The fill weight of a certain brand of adult cereal is normally distributed with a mean of 910 grams and a standard deviation of 5 grams. If we select one box of cereal at random from this population, what is the probability that it will weigh more than 904 grams? (Ch6) A. 0.3849 B. 0.8849 C. 0.1151 D. 0.7698 E. 0.2302 8. Consider a sampling distribution formed based on a small sample of 5 items only. The standard deviation of the sampling distribution of all sample means (sigma sub X-bar) is ______________ ______________ than the standard deviation of the population of individual measurements. (Ch7) A. always less B. sometimes less C. always more D. sometimes more E. can be equal to or less than or greater 9. If the sampled population has a mean 48 and standard deviation 16, then the mean and the standard deviation for the sampling distribution of the sample mean X-bar for n=64 are: (Ch7) A. 4 and 4 B. 12 and 4 C. 48 and 2 D. 48 and ¼ E. 48 and 16 10. If the wages of workers for a given company are normally distributed with a mean of $15 per hour, then the proportion of the workers earning more than $13 per hour: (Ch6) A. Is greater than the proportion earning less than $18 per hour B. Is less than 50% C. Is less than the proportion earning more than the mean wage D. Is greater than the proportion earning less than $13 per hour E. Cannot tell without knowing the standard deviation 11. In a manufacturing process, we are interested in measuring the average length of a certain type of bolt. Past data indicates that the standard deviation is .25 inches. How many bolts should be sampled in order to make us 95% confident that the sample mean bolt length is within .02 inches of the mean bolt length? (Ch8) A. 25 B. 49 C. 423 D. 601 E. 1225 12. In a study of factors affecting soldiers' decisions to reenlist, 64 subjects were measured for an index of satisfaction and the sample mean is 60 and the sample standard deviation is 16. Use the given sample data to construct the 99% confidence interval for the population mean. (Ch8) A. 56.00 to 64.00 B. 56.66 to 63.34 C. 58.00 to 62.00 D. 53.20 to 66.80 E. 54.69 to 65.31 13. Consider two population distributions labeled X and Y. Distribution X is highly skewed while the distribution Y is slightly skewed. In order for the sampling distributions of X and Y to achieve the same degree of normality: (Ch7) A. Population Y will require a larger sample size B. Population X will require a larger sample size C. Population X and Y will require the same sample size D. None of the above 14. If a population distribution is known to be normal, then it follows that: (Ch 7) A. The sample mean is equal to the population mean B. The distribution of sample mean is skewed for small samples but becomes more and more normal as the sample size increases C. The sample standard deviation equals the population standard deviation D. The sample proportion equals the population proportion E. None of the above 15. A continuity correction in normal approximation of a binomial distribution is not needed if:(Ch6) A. np>=10 B. n(1-p)>= 10 C. np(1-p)>= 10 D. np>=10 and n(1-p)>=10 16. In a manufacturing process a random sample of 9 bolts is taken which gives a mean length of 3 inches with a variance of .09. What is the 90% confidence interval for the mean length of the bolt? (Ch8) A. 2.8355 to 3.1645 B. 2.5065 to 3.4935 C. 2.4420 to 3.5580 D. 2.8140 to 3.1860 E. 2.9442 to 3.0558 Essay Type Question (each 12 points) 1. The following frequency table summarizes the distances in miles of 70 patients from a regional hospital. Distance Frequency 0-2 25 2-4 30 4-6 10 6-8 5 Calculate the sample standard deviation for this data (since it is a case of grouped data- use group or class midpoints in the formula in place of X values). (Ch3) 2. Container 1 has 10 items, 3 of which are defective. Container 2 has 6 items, 2 of which are defective. If one item is drawn independently from each container find the probability distribution for X defined as the number of defective items drawn (Hint: You have to use both multiplicative and additive rules to find P(X=1), whereas P(X=0) and P(X=2) can be found only by multiplicative rules). (Ch4) 3. The weight of a product is normally distributed with a standard deviation of .5 ounces. What should the average weight be if the production manager wants no more than 10% of the products to weigh more than 5.8 ounces? (Ch6) 4. The number of items rejected daily by a manufacturer because of defects for the last 30 days are: 20, 21, 8, 17, 22, 19, 18, 19, 14, 17, 11, 6, 21, 25, 4, 18, 8, 12, 16, 16, 10, 28, 24, 6, 21, 20, 23, 5, 17, 8 . Complete this frequency table for the above data (note that the class intervals exclude the upper boundaries): (Ch2) Class Frequency Rel. Freq Cum. Rel. Freq 4<9 9<14 14<19 19<24 24<29 iv) RF(19<24) v) RF(24<29) 5. The probability that an appliance is in repair is .6. If an apartment complex has 100 such appliances, what is the probability that at least 70 will be in repair? Use the normal approximation to the binomial and show the Z values and the steps used in the calculation. (Ch6) 6. An insurance company estimates 45 percent of its claims have errors. The insurance company wants to estimate with 90 percent confidence the proportion of claims with errors. What sample size is needed if they wish to be within 5 percentage points of the actual? (Ch8) 7. In a local survey, 100 citizens indicated their opinions on a revision to a local land use plan. Of the 62 favorable responses, there were 40 males. Of the 38 unfavorable responses, there were 15 males. If one citizen is randomly selected find the probability of “a male or has an unfavorable opinion” (Hint: It would be easy to solve this problem if you first build the contingency table of frequencies with Male/Female and Yes/No in rows and columns and use rule of addition) (Ch4) 8. It has been reported that the average time to download the home page from a government website was 0.9 seconds. Suppose that the download times were normally distributed with a standard deviation of 0.3 seconds. If random samples of 36 download times are selected what two values symmetrically distributed around the population mean have 80% probability of containing the sample mean? (Ch8) 9. A small town has a population of 15,000 people. Among these 1,000 regularly visit a popular local bar. A sample of 121 people who visit the bar is surveyed for their annual expenditures in the bar. It is found that on average each person who regularly visits the bar spends about $2500 per year in the bar with a standard deviation of $196. Construct a 95 percent confidence interval around the mean annual expenditure in the bar. (Chs. 7 and 8) [Show More]

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