Q&A Hero

Q: The margin of error is: A) the largest possible sampling error at a specified level of confidence. B) the critical value multiplied by the standard error of the sampling distribution. C) Both A and B D) the difference between the point estimate and the parameter.

Q: In an effort to estimate the mean dollars spent per visit by customers of a food store, the manager has selected a random sample of 100 cash register receipts. The mean of these was $45.67 with a sample standard deviation equal to $12.30. Assuming that he wants to develop a 90 percent confidence interval estimate, the upper limit of the confidence interval estimate is: A) about $2.02 B) approximately $65.90 C) about $47.69 D) None of the above

Q: Which of the following will increase the width of a confidence interval (assuming that everything else remains constant)? A) Decreasing the confidence level B) Increasing the sample size C) A decrease in the standard deviation D) Decreasing the sample size

Q: In developing a confidence interval estimate for the population mean, which of the following is true? A) The larger the sample standard deviation, the wider will be the interval estimate, all other things being equal. B) If the population standard deviation is unknown, the appropriate critical value should be obtained from the t-distribution. C) The confidence interval developed from a smaller sample size will have a larger margin of error than one obtained using a larger sample size, all other things being equal. D) All of the above are true.

Q: In an application to estimate the mean number of miles that downtown employees commute to work roundtrip each day, the following information is given:Based on this information, the upper limit for a 95 percent confidence interval estimate for the true population mean is:A) about 5.97 miles.B) about 7.83 miles.C) nearly 12.0 miles.D) about 5.86 miles.

Q: The general format for a confidence interval is: A) point estimate z (standard deviation). B) point estimate (critical value)(standard error). C) margin of error (confidence coefficient) (standard error). D) point estimate (critical value)(standard deviation)

Q: Sampling error occurs when: A) a nonstatistical sample is used. B) the statistic computed from the sample is not equal to the parameter for the population. C) a random sample is used rather than a convenience sample. D) a confidence interval is used to estimate a population value rather than a point estimate.

Q: Which of the following statements applies to a point estimate? A) The point estimate is a parameter. B) The point estimate will tend to be accurate if the sample size exceeds 30 for non-normal populations. C) The point estimate is subject to sampling error and will almost always be different from the population value. D) The point estimate is needed to determine the required sample size when estimating the population mean.

Q: Twenty one (21) customers out of a simple random sample of 50 said that they came to the grocery store within one month after getting a card from the company. Based on the data, the marketing manager thinks that the number of total customers who would come in to the grocery store within one month is 4,200 because 10,000 cards were mailed.

Q: When determining sample size for a proportion, the farther that p is from 0.5, the smaller the resulting sample size will be.

Q: Chicago Connection, a local pizza company, delivers pizzas for free within the market area. The delivery drivers are paid $2.00 per delivery plus they get to keep any tips. To estimate the proportion of deliveries that result in a tip to the driver, a random sample of 64 deliveries was selected. Of these, 48 times a tip was received. Based on this information, and using a 95 percent confidence level, the margin of error for the estimate is approximately .1061.

Q: The manager of the local county fair believes that no more than 30 percent of the adults in the county would object to a fee increase to attend the fair if it meant that better entertainment could be secured. To estimate the true proportion, he has selected a random sample of 200 adults. The manager will use a 90 percent confidence level. Assuming his assumption about the 30 percent holds, the margin of error for the estimate will be approximately .169.

Q: In determining the sample size requirement for an application involving the estimation of the proportion of department store customers who pay using the store's credit card, the closer the true proportion is to .5, the larger will be the required sample size for a given margin of error and confidence level.

Q: A random sample of n = 500 people was surveyed recently to determine an estimate for the proportion of people in the population who had attended at least some college. The estimate concluded that between 0.357 and 0.443 of the population had attended. Given this information, we can determine that the confidence level was approximately 95 percent.

Q: Recently, a report in a financial journal indicated that the 90 percent confidence interval estimate for the proportion of investors who own one or more mutual funds is between 0.88 and 0.92. Given this information, the sample size that was used in this study was approximately 609 investors.

Q: The t-distribution is used for the critical value when estimating a population proportion when the standard deviation of the population is not known.

Q: When determining sample size for a proportion, using p = 0.5 will produce the smallest possible value for n.

Q: A publisher is interested in estimating the proportion of textbooks that students resell at the end of the semester. He is interested in making this estimate using a confidence level of 95 percent and a margin of error of 0.02. Based upon his prior experience, he believes that π is somewhere around 0.60. Given this information, the required sample size is over 2,300 students.

Q: A local pizza company is interested in estimating the percentage of customers who would take advantage of a coupon offer. To do this, they give the coupon out to a random sample of 100 customers. Of these, 45 actually use the coupon. Based on a 95 percent confidence level, the upper and lower confidence interval limits are approximately 0.3525 to 0.5475.

Q: The procurement manager for a large company wishes to estimate the proportion of parts from a supplier that are defective. She has selected a random sample of n = 200 incoming parts and has found 11 to be defective. Based on a 95 percent confidence level, the upper and lower limits for the confidence interval estimate are approximately 0.0234 to 0.0866.

Q: For a given sample size and a given confidence level, the closer p is to 1.0, the greater the margin of error will be.

Q: In estimating a population proportion, the factors that are needed to determine the required sample size are the confidence level, the margin of error and some idea of what the population proportion is.

Q: When determining the sample size for a proportion, if you have no previous information available to estimate p, then the best value to use is π = 0.5.

Q: The one sure thing that can be said about a point estimate is that it will most likely be subject to sampling error and not equal the parameter.

Q: A Parks and Recreation official surveyed 200 people at random who have used one of the city's parks. The survey revealed that 26 resided outside the city limits. If she had to arrive at one single value to estimate the true proportion of park users who are residents of the city it would be 0.13.

Q: The concept of margin of error applies directly when estimating a population mean, but is not appropriate when estimating a population proportion.

Q: In determining the required sample size when estimating a population proportion, it is necessary to start with some idea of what that proportion is.

Q: When σ is unknown, one should use an estimate that is considered to be at least as large as the true σ. This will provide a conservative sample size to be on the safe side.

Q: A random sample of 100 people was selected from a population of customers at a local bank. The mean age of these customers was 40. If the population standard deviation is thought to be 5 years, the margin of error for a 95 percent confidence interval estimate is .98 year.

Q: A grocery store manager is interested in estimating the mean weight of apples received in a shipment. If she wishes to have the estimate be within .05 pound with 90 percent confidence, the sample size should be 103 apples if she believes that the standard deviation is .08 pound.

Q: In estimating a population mean, increasing the confidence level will result in a higher margin of error for a given sample size.

Q: An emergency room in a hospital wants to determine the sample size needed for estimating their mean number of visits per day. If they want a 99 percent confidence level the correct critical value to use is 2.33.

Q: After taking a speed-reading course, students are supposed to be able to read faster than they could before taking the course. A pilot sample of n = 25 students showed a mean increase of 300 words per minute with a standard deviation equal to 60 words per minute. To estimate the population mean with 95 percent confidence and a margin of error of 10 minutes, the required sample size is approximately 139 students.

Q: The State Transportation Department wishes to estimate the mean speed of vehicles on a certain stretch of highway. They wish to estimate the mean within 0.75 mph and they wish to have a confidence level equal to 99 percent. Based on this information only, they can determine that the required sample size is 320 vehicles.

Q: When a decision maker determines the required sample size for estimating a population mean, a change in the confidence level will result in a change in the required sample size, provided that the margin of error is also modified accordingly.

Q: An analyst for a financial investment firm recently went through the effort to determine the required sample size for estimating the mean number of transactions per year for the clients of his firm. The calculations, which were based on a 95 percent confidence level and a margin of error of 3, gave a required sample size of 300. Given this information, the value used for the population standard deviation must have been about 26.5 transactions.

Q: A university computer lab manager wishes to estimate the mean time that students stay in the lab per visit. She believes that the population standard deviation would be no larger than 10 minutes. Further, she wishes to have a confidence level of 90 percent and a margin of error of 2.00 minutes. Given this, the sample size that she uses is approximately 60 students.

Q: The manager in charge of concessions at an NFL football stadium is interested in estimating the mean dollars that are spent per person attending the games. A pilot sample of n = 50 people has revealed a sample mean and standard deviation of $12.35 and $2.35 respectively. He wishes to estimate the population mean within $0.20 of the true mean and wishes to have a confidence level of 95 percent. Given this, he needs to sample an additional 481 people.

Q: In a sample size determination situation, reducing the margin of error by half will double the required sample size.

Q: In an effort to estimate the mean length of stay for motel guests at a major national motel chain, the decision makers asked for a 95 percent confidence, and a margin of error of 0.25 days. Further, it was known that the population standard deviation is 0.50 days. Given this, the required sample size to estimate the mean length of stay is about 16 customers.

Q: If a decision maker desires a small margin of error and a high level of confidence, it is certain that the required sample size will be quite large.

Q: When finding sample size, cutting the margin of error in half requires that the sample size be four times larger.

Q: If a pilot sample of n = 40 items has been used as a first step in determining a required sample size of n = 360, the decision maker can go ahead and use these 40 and take a sample of only 320 more items.

Q: The national sales manager for a textbook publishing company wishes to estimate the mean number of books sold per college. She wishes to have her estimate be within 30 copies and wants a 95 percent confidence interval estimate. If a pilot sample of 30 schools gave a sample standard deviation equal to 60 books, the required total sample size is less than the pilot sample already taken.

Q: The main purpose of a pilot sample in an application involving an estimate for a population mean is to determine what the margin of error will likely be.

Q: In determining the required sample size in an application involving an estimate for the population mean, if the population standard deviation is known, there is no compelling reason to select a pilot sample.

Q: Two confidence interval estimates were developed from the same sample of a population. The wider interval will be the one that has the higher confidence level.

Q: When estimating sample size, a 90 percent confidence level will result in a smaller sample size than a 95 percent confidence level.

Q: A bank manager wishes to estimate the mean waiting time spent by customers at his bank. He knows from previous experience that the standard deviation is about 4.0 minutes. If he desires a 90 percent confidence interval estimate and wishes to have a margin of error of 1 minute, the required sample size will be approximately 143.

Q: All of the factors that are needed to determine the required sample size are within the control of the decision maker.

Q: One factor that plays an important part in determining what the needed sample size is when developing a confidence interval estimate is the level of confidence that you wish to use.

Q: A pilot sample is one that is used when a decision maker wishes to get an advance idea of what the mean of the population might be.

Q: In estimating a population mean, a large sample size is generally preferable to a small sample size because the margin of error is generally smaller.

Q: If the population is not normally distributed, the t-distribution cannot be used.

Q: All other factors held constant, the higher the confidence level, the closer the point estimate for the population mean will be to the true population mean.

Q: A 95 percent confidence interval estimate will have a margin of error that is approximately 95 percent of the size of the population mean.

Q: The impact on the margin of error for a confidence interval for an increase in confidence level and a decrease in sample size is unknown since these changes are contradictory.

Q: The t-distribution is used to obtain the critical value in developing a confidence interval when the population distribution is not known or the sample size is small.

Q: The makers of weight loss product are interested in estimating the mean weight loss for users of their product. To do this, they have selected a random sample of n = 9 people and have provided them with a supply of the product. After six months, the nine people had an average weight loss of 15.3 pounds with a standard deviation equal to 3.5 pounds. The upper limit for the 90 percent confidence interval estimate for the population mean is approximately 17.47 pounds.

Q: A 95 percent confidence interval for a mean will contain 95 percent of the population within the interval.

Q: The bottlers of a new fruit juice daily select a random sample of 12 bottles of the drink to estimate the mean quantity of juice in the bottles filled that day. On one such day, the following results were observed: = 12.03; s = 0.12. Based on this information, the margin of error associated with a 90 percent confidence interval estimate for the population mean is 1.7959 ounces.

Q: How would you respond to a statement that says that by increasing the sample size, the amount of sampling error will be decreased?

Q: Explain what is meant by the concept of sampling distribution.

Q: According to the most recent Labor Department data, 10.5% of engineers (electrical, mechanical, civil, and industrial) were women. Suppose a random sample of 50 engineers is selected. How likely is it that the random sample will contain fewer than 5 women in these positions? A) 0.4522 B) 0.3124 C) 0.5121 D) 0.5512

Q: The National Association of Realtors released a survey indicating that a surprising 43% of first-time home buyers purchased their homes with no-money-down loans during 2005. The fear is that house prices will decline and leave homeowners owing more than their homes are worth. PMI Mortgage Insurance estimated that there existed a 50% risk that prices would decline within two years in major metro areas such as San Diego, Boston, Long Island, New York City, Los Angeles, and San Francisco. A survey taken by realtors in the San Francisco area found that 12 out of the 20 first-time home buyers sampled purchased their home with no-money-down loans. Calculate the probability that at least 12 in a sample of 20 first-time buyers would take out no-money-down loans if San Francisco's proportion is the same as the nationwide proportion of no-money-down loans. A) 0.0618 B) 0.0124 C) 0.0512 D) 0.0441

Q: The J R Simplot Company is one of the world's largest privately held agricultural companies, employing over 10,000 people in the United States, Canada, China, Mexico, and Australia. More information can be found at the company's Web site: www.Simplot.com. One of its major products is french fries that are sold primarily on the commercial market to customers such as McDonald's and Burger King. French fries have numerous quality attributes that are important to customers. One of these is called "dark ends," which are the dark-colored ends that can occur when the fries are cooked. Suppose a major customer will accept no more than 0.06 of the fries having dark ends. Recently, the customer called the Simplot Company saying that a recent random sample of 300 fries was tested from a shipment and 27 fries had dark ends. Assuming that the population does meet the 0.06 standard, what is the probability of getting a sample of 300 with 27 or more dark ends? A) 0.0341 B) 0.0162 C) 0.0012 D) 0.0231

Q: United Manufacturing and Supply makes sprinkler valves for use in residential sprinkler systems. United supplies these valves to major companies such as Rain Bird and Nelson, who in turn sell sprinkler products to retailers. United recently entered into a contract to supply 40,000 sprinkler valves. The contract called for at least 97% of the valves to be free of defects. Before shipping the valves, United managers tested 200 randomly selected valves and found 190 defect-free valves in the sample. The managers wish to know the probability of finding 190 or fewer defect-free valves if in fact the population of 40,000 valves is 97% defect-free. The probability is: A) 0.0111 B) 0.0612 C) 0.0475 D) 0.0212

Q: Given a population in which the probability of success is p = 0.20, if a sample of 500 items is taken, then calculate the probability the proportion of successes in the sample will be between 0.18 and 0.23 if the sample size is 200. A) 0.8911 B) 0.7121 C) 0.8712 D) 0.6165

Q: Given a population where the proportion of items with a desired attribute is p = 0.25, if a sample of 400 is taken, what is the probability the proportion of successes in the sample will be greater than 0.22? A) 0.9162 B) 0.8812 C) 0.7141 D) 0.8412

Q: The proportion of items in a population that possess a specific attribute is known to be 0.70. If a simple random sample of size n = 100 is selected and the proportion of items in the sample that contain the attribute of interest is 0.65, what is the sampling error? A) -0.03 B) -0.05 C) 0.08 D) 0.01

Q: If a random sample of 200 items is taken from a population in which the proportion of items having a desired attribute is p = 0.30, what is the probability that the proportion of successes in the sample will be less than or equal to 0.27? A) 0.0841 B) 0.1011 C) 0.1912 D) 0.1762

Q: A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find P(≥ 0.27). A) 0.7422 B) 0.8141 C) 0.6125 D) 0.6841

Q: In an article entitled "Fuel Economy Calculations to Be Altered," James R. Healey indicated that the government planned to change how it calculates fuel economy for new cars and trucks. This is the first modification since 1985. It is expected to lower average mileage for city driving in conventional cars from 10% to 20%. AAA has forecast that the 2008 Ford F-150 would achieve 15.7 mile per gallon (mpg). The 2008 Ford F-150 was tested by AAA members driving the vehicle themselves and was found to have an average of 14.3 mpg. Assume that the mean obtained by AAA members is the true mean for the population of 2008 Ford F-150 trucks and that the population standard deviation is 5 mpg. The current method of calculating the mpg forecasts that the 2008 F-150 will average 16.8 mpg. Determine the probability that these same 100 AAA members would average more than 16.8 mpg while testing the 2008 F-150. A) 0 B) 0.0155 C) 0.0412 D) None of the above

Q: The branch manager for United Savings and Loan in Seaside, Virginia, has worked with her employees in an effort to reduce the waiting time for customers at the bank. Recently, she and the team concluded that average waiting time is now down to 3.5 minutes with a standard deviation equal to 1.0 minute. However, before making a statement at a managers' meeting, this branch manager wanted to double-check that the process was working as thought. To make this check, she randomly sampled 25 customers and recorded the time they had to wait. She discovered that mean wait time for this sample of customers was 4.2 minutes. Based on the team's claims about waiting time, what is the probability that a sample mean for n = 25 people would be as large or larger than 4.2 minutes? A) 0.0214 B) 0.0512 C) 0.0231 D) 0.0011

Q: Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. If the manufacturer of the battery is able to reduce the standard deviation of battery life from 10 to 9 hours, what would be the probability that 16 batteries randomly sampled from the population will have a sample mean life of between 70 and 80 hours?A) 0.6127B) 0.8124C) 0.9736D) 0.8812

Q: SeeClear Windows makes windows for use in homes and commercial buildings. The standards for glass thickness call for the glass to average 0.375 inches with a standard deviation equal to 0.050 inch. Suppose a random sample of n = 50 windows yields a sample mean of 0.392 inches. What is the probability if the windows meet the standards? A) 0.0612 B) 0.0082 C) 0.0015 D) 0.0009

Q: A random sample of 100 items is selected from a population of size 350. What is the probability that the sample mean will exceed 200 if the population mean is 195 and the population standard deviation equals 20? (Hint: Use the finite correction factor since the sample size is more than 5% of the population size.) A) 0.0415 B) 0.0016 C) 0.0241 D) 0.0171

Q: Suppose nine items are randomly sampled from a normally distributed population with a mean of 100 and a standard deviation of 20. The nine randomly sampled values are: 125 95 66 116 99 91 102 51 110 Calculate the probability of getting a sample mean that is smaller than the sample mean for these nine sampled values. A) 0.1411 B) 0.1612 C) 0.1512 D) 0.2266

Q: A normally distributed population has a mean of 500 and a standard deviation of 60. Determine the probability that a random sample of size 25 selected from the population will have a sample mean greater than or equal to 515. A) 0.1056 B) 0.1761 C) 0.0712 D) 0.0151

Q: Suppose that a population is known to be normally distributed with mean = 2,000 and standard deviation = 230. If a random sample of size n = 8 is selected, calculate the probability that the sample mean will exceed 2,100.A) 0.2141B) 0.1871C) 0.0712D) 0.1093