Business Communication

Q: A major issue facing many states is whether to legalize casino gambling. Suppose the governor of one state believes that more than 55% of the state's registered voters would favor some form of legal casino gambling. However, before backing a proposal to allow such gambling, the governor has instructed his aides to conduct a statistical test on the issue. To do this, the aides have hired a consulting firm to survey a simple random sample of 300 voters in the state. Of these 300 voters, 175 actually favored legalized gambling.State the appropriate null and alternative hypotheses.A) H0 : p = 0.58 Ha: p 0.58B) H0 : p 0.55 Ha: p > 0.55C) H0 : p = 0.55 Ha: p 0.55D) H0 : p 0.58 Ha : p > 0.58

Q: An issue that faces individuals investing for retirement is allocating assets among different investment choices. Suppose a study conducted 10 years ago showed that 65% of investors preferred stocks to real estate as an investment. In a recent random sample of 900 investors, 540 preferred real estate to stocks. Is this new data sufficient to allow you to conclude that the proportion of investors preferring stocks to real estate has declined from 10 years ago? Conduct your analysis at the = 0.02 level of significance.A) Because z = -1.915 is not less than -2.055, do not reject H0. A higher proportion of investors prefer stocks today than 10 years ago.B) Because z = -1.915 is not less than -2.055, do not reject H0. A lower proportion of investors prefer stocks today than 10 years ago.C) Because z = -3.145 is less than -2.055, reject H0. A lower proportion of investors prefer stocks today than 10 years ago.D) Because z = -3.145 is less than -2.055, reject H0. A higher proportion of investors prefer stocks today than 10 years ago.

Q: Suppose a recent random sample of employees nationwide that have a 401(k) retirement plan found that 18% of them had borrowed against it in the last year. A random sample of 100 employees from a local company who have a 401(k) retirement plan found that 14 had borrowed from their plan. Based on the sample results, is it possible to conclude, at the = 0.025 level of significance, that the local company had a lower proportion of borrowers from its 401(k) retirement plan than the 18% reported nationwide?A) The z-critical value for this lower tailed test is z = -1.96. Because -1.5430 is greater than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average.B) The z-critical value for this lower tailed test is z = -1.96. Because -1.0412 is greater than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average.C) The z-critical value for this lower tailed test is z = 1.96. Because 1.5430 is less than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average.D) The z-critical value for this lower tailed test is z = 1.96. Because 1.0412 is less than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average.

Q: For the following hypothesis test With n = 100 and p = 0.66, state the conclusion. A) Because the computed value of z = -2.0785 is less than the critical value of z = -1.96, reject the null hypothesis and conclude that the population proportion is less than 0.75. B) Because the computed value of z = -0.3412 is less than the critical value of z = -1.645, reject the null hypothesis and conclude that the population proportion is less than 0.75. C) Because the computed value of z = 1.4919 is greater than the critical value of z = -1.96, accept the null hypothesis and conclude that the population proportion is greater than 0.75. D) Because the computed value of z = -0.3412 is greater than the critical value of z = -1.645, accept the null hypothesis and conclude that the population proportion is greater than 0.75.

Q: For the following hypothesis test With n = 100 and p = 0.66, state the calculated value of the test statistic. A) 2.7299 B) -2.0785 C) 1.4919 D) -0.3421

Q: For the following hypothesis test: With n= 0.42 and p = 0.42, state the conclusion A) Because the calculated value of the test statistic, t=0.4122, is neither greater than 2.013 nor less than -2.013, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. B) Because the calculated value of the test statistic, t=1.7291, is neither greater than 2.013 nor less than -2.013, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. C) Because the calculated value of the test statistic, z = 1.2412, is neither greater than 2.575 nor less than -2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. D) Because the calculated value of the test statistic, z = 0.3266, is neither greater than 2.575 nor less than -2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40.

Q: For the following hypothesis test: With n = 64 and p = 0.42, state the calculated value of the test statistic A) t = 0.4122 B) t = 1.7291 C) z = 0.3266 D) z = 1.2412

Q: For the following hypothesis test: With n= 64 and p= 0.42, state the decision rule in terms of the critical value of the test statistic A) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than 2.013 or less than -2.013. Otherwise, do not reject. B) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than 2.013 or greater than -2.013. Otherwise, do not reject. C) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than 2.575 or less than -2.575. Otherwise, do not reject. D) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than 2.575 or greater than -2.575. Otherwise, do not reject.

Q: Given the following null and alternativeTest the hypothesis using = 0.01 assuming that a sample of n = 200 yielded x = 105 items with the desired attribute.A) Since -2.17 > -2.33, the null hypothesis is not rejected.B) Since -1.86 > -1.02, the null hypothesis is not rejected.C) Since -2.17 > -2.33, the null hypothesis is rejected.D) Since -1.86 > -1.02, the null hypothesis is rejected.

Q: Hono Golf is a manufacturer of golf products in Taiwan and China. One of the golf accessories it produces at its plant in Tainan Hsing, Taiwan, is plastic golf tees. The injector molder produces golf tees that are designed to have an average height of 66 mm. To determine if this specification is met, random samples are taken from the production floor. One sample is contained in the file labeled THeight. If the hypothesis test determines the specification is not being met, the production process will be shut down while causes and remedies are determined. At times this occurs even though the process is functioning to specification. What type of statistical error would this be? A) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type II error. B) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type I error. C) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type II error. D) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type I error.

Q: Hono Golf is a manufacturer of golf products in Taiwan and China. One of the golf accessories it produces at its plant in Tainan Hsing, Taiwan, is plastic golf tees. The injector molder produces golf tees that are designed to have an average height of 66 mm. To determine if this specification is met, random samples are taken from the production floor. One sample is contained in the file labeled THeight. Determine if the process is not producing the tees to specification. Use a significance level of 0.01. A) Since t = 2.1953 < 2.8073 do not reject H0. There is not sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. B) Since t = 2.1953 < 2.8073 reject H0. There is sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. C) Since t = 1.2814 < 1.9211 do not reject H0. There is not sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. D) Since t = 1.2814 < 1.9211 reject H0. There is sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm.

Q: The Center on Budget and Policy Priorities (www.cbpp.org) reported that average out-of-pocket medical expenses for prescription drugs for privately insured adults with incomes over 200% of the poverty level was $173 in 2002. Suppose an investigation was conducted in 2012 to determine whether the increased availability of generic drugs, Internet prescription drug purchases, and cost controls have reduced out-of-pocket drug expenses. The investigation randomly sampled 196 privately insured adults with incomes over 200% of the poverty level, and the respondents' 2012 out-of-pocket medical expenses for prescription drugs were recorded. These data are in the file Drug Expenses. Based on the sample data, can it be concluded that 2012 out-of-pocket prescription drug expenses are lower than the 2002 average reported by the Center on Budget and Policy Priorities? Use a level of significance of 0.01 to conduct the hypothesis test. A) Because t = -2.69 is less than -2.3456, do not reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average. B) Because t = -2.69 is less than -2.3456, reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average. C) Because t = -1.69 is less than -0.8712, reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average. D) Because t = -1.69 is less than -0.8712, do not reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average.

Q: The U.S. Bureau of Labor Statistics (www.bls.gov) released its Consumer Expenditures report in October 2008. Among its findings is that average annual household spending on food at home was $3,624. Suppose a random sample of 137 households in Detroit was taken to determine whether the average annual expenditure on food at home was less for consumer units in Detroit than in the nation as a whole. The sample results are in the file Detroit Eats. Based on the sample results, can it be concluded at the = 0.02 level of significance that average consumer-unit spending for food at home in Detroit is less than the national average?A) Because t = -13.2314 is less than the critical t value of -1.4126, do not reject H0.The annual average consumer unit spending for food at home in Detroit is not less than the 2006 national consumer unit averageB) Because t = -13.2314 is less than the critical t value of -1.4126, reject H0.The annual average consumer unit spending for food at home in Detroit is less than the 2006 national consumer unit averageC) Because t = -15.7648 is less than the critical t value of -2.0736, do not reject H0.The annual average consumer unit spending for food at home in Detroit is not less than the 2006 national consumer unit average.D) Because t = -15.7648 is less than the critical t value of -2.0736, reject H0.The annual average consumer unit spending for food at home in Detroit is less than the 2006 national consumer unit average.

Q: At a recent meeting, the manager of a national call center for a major Internet bank made the statement that the average past-due amount for customers who have been called previously about their bills is now no larger than $20.00. Other bank managers at the meeting suggested that this statement may be in error and that it might be worthwhile to conduct a test to see if there is statistical support for the call center manager's statement. The file called Bank Call Center contains data for a random sample of 67 customers from the call center population. Assuming that the population standard deviation for past due amounts is known to be $60.00, what should be concluded based on the sample data? Test using = 0.10.A) Because p-value = 0.4121 > alpha = 0.10, we do not reject the null hypothesis.The sample data do not provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less.B) Because p-value = 0.4121 > alpha = 0.10, we reject the null hypothesis.The sample data provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less.C) Because p-value = 0.2546 > alpha = 0.10, we do not reject the null hypothesis.The sample data do not provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less.D) Because p-value = 0.2546 > alpha = 0.10, we reject the null hypothesis.The sample data provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less.

Q: The makers of Mini-Oats Cereal have an automated packaging machine that can be set at any targeted fill level between 12 and 32 ounces. Every box of cereal is not expected to contain exactly the targeted weight, but the average of all boxes filled should. At the end of every shift (eight hours), 16 boxes are selected at random and the mean and standard deviation of the sample are computed. Based on these sample results, the production control manager determines whether the filling machine needs to be readjusted or whether it remains all right to operate. At the end of a particular shift during which the machine was filling 24-ounce boxes of Mini-Oats, the sample mean of 16 boxes was 24.32 ounces, with a standard deviation of 0.70 ounce. Assist the production control manager in determining if the machine is achieving its targeted average using test statistic and critical value t. Conduct the test using a p-value. A) p-value = 0.0872 > 0.025; therefore do not reject H0 B) p-value = 0.0422 > 0.005; therefore do not reject H0 C) p-value = 0.0314 < 0.105; therefore reject H0 D) p-value = 0.0121< 0. 0805; therefore reject H0

Q: The makers of Mini-Oats Cereal have an automated packaging machine that can be set at any targeted fill level between 12 and 32 ounces. Every box of cereal is not expected to contain exactly the targeted weight, but the average of all boxes filled should. At the end of every shift (eight hours), 16 boxes are selected at random and the mean and standard deviation of the sample are computed. Based on these sample results, the production control manager determines whether the filling machine needs to be readjusted or whether it remains all right to operate. Use = 0.05. At the end of a particular shift during which the machine was filling 24-ounce boxes of Mini-Oats, the sample mean of 16 boxes was 24.32 ounces, with a standard deviation of 0.70 ounce. Assist the production control manager in determining if the machine is achieving its targeted average using test statistic and critical value t.A) Since -1.2445 < 1.013 < 1.2445, do not reject H0 and conclude that the filling machine remains all right to operate.B) Since -1.2445 < 1.013 < 1.2445, reject H0 and conclude that the filling machine needs to be moderated.C) Since -2.1315 < 1.83 < 2.1315, do not reject H0 and conclude that the filling machine remains all right to operate.D) Since -2.1315 < 1.83 < 2.1315, reject H0 and conclude that the filling machine needs to be moderated.

Q: A mail-order business prides itself in its ability to fill customers' orders in six calendar days or less on the average. Periodically, the operations manager selects a random sample of customer orders and determines the number of days required to fill the orders. Based on this sample information, he decides if the desired standard is not being met. He will assume that the average number of days to fill customers' orders is six or less unless the data suggest strongly otherwise. On one occasion where a sample of 40 customers was selected, the average number of days was 6.65, with a sample standard deviation of 1.5 days. Can the operations manager conclude that his mail-order business is achieving its goal? Use a significance level of 0.025 to answer this question. Conduct the test using this p-value. A) Since 0.024 > 0.0041, reject the null hypothesis. B) Since 0.046 > 0.0025, reject the null hypothesis. C) Since 0.0046 < 0.025, reject the null hypothesis. D) Since 0.0024 < 0.041, reject the null hypothesis.

Q: A mail-order business prides itself in its ability to fill customers' orders in six calendar days or less on the average. Periodically, the operations manager selects a random sample of customer orders and determines the number of days required to fill the orders. Based on this sample information, he decides if the desired standard is not being met. He will assume that the average number of days to fill customers' orders is six or less unless the data suggest strongly otherwise. On one occasion where a sample of 40 customers was selected, the average number of days was 6.65, with a sample standard deviation of 1.5 days. Can the operations manager conclude that his mail-order business is achieving its goal? Use a significance level of 0.025 to answer this question. A) Since 2.2216 < 2.4511, reject H0 and conclude that the mail-order business is not achieving its goal B) Since 2.7406 > 2.023, reject H0 and conclude that the mail-order business is not achieving its goal. C) Since 2.4421 > 2.023, reject H0 and conclude that the mail-order business is not achieving its goal. D) Since 2.2346 < 2.5113, reject H0 and conclude that the mail-order business is not achieving its goal.

Q: The director of a state agency believes that the average starting salary for clerical employees in the state is less than $30,000 per year. To test her hypothesis, she has collected a simple random sample of 100 starting clerical salaries from across the state and found that the sample mean is $29,750. Assuming the population standard deviation is known to be $2,500 and the significance level for the test is to be 0.05, what is the critical value (stated in dollars)? A) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.645. Solving for the critical x-bar: -1.645 = (x-bar - 30,000)/250, x-bar = $29,588.75 B) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.96. Solving for the critical x-bar: -1.96 = (x-bar - 30,000)/250, x-bar = $34,211.14 C) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.645. Solving for the critical x-bar: -1.645 = (x-bar - 30,000)/250, x-bar = $34,211.14 D) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.96. Solving for the critical x-bar: -1.96 = (x-bar - 30,000)/250, x-bar = $30,411.25

Q: The National Club Association does periodic studies on issues important to its membership. The 2012 Executive Summary of the Club Managers Association of America reported that the average country club initiation fee was $31,912. Suppose a random sample taken in 2009 of 12 country clubs produced the following initiation fees: $29,121$31,472$28,054$31,005$36,295$32,771$26,205$33,299$25,602$33,726$39,731$27,816Based on the sample information, can you conclude at the = 0.05 level of significance that the average 2009 country club initiation fees are lower than the 2008 average? Conduct your test at the level of significance.A) Because t = -0.4324 is not less than t critical = -1.4512, do not reject Ho. The 2009 average country club initiation fee is not less than the 2008 average.B) Because t = -0.4324 is not less than t critical = -1.4512, reject Ho. The 2009 average country club initiation fee is less than the 2008 average.C) Because t = -0.5394 is not less than t critical = -1.7959, do not reject Ho. The 2009 average country club initiation fee is not less than the 2008 average.D) Because t = -0.5394 is not less than t critical = -1.7959, reject Ho. The 2009 average country club initiation fee is less than the 2008 average.

Q: For the following hypothesis:With n = 20, = 71.2, s = 6.9, and = 0.1, state the conclusion.A) Because the computed value of t = 0.78 is not greater than 2.1727, reject the null hypothesis.B) Because the computed value of t = 0.78 is not greater than 2.1727, do not reject the null hypothesis.C) Because the computed value of t = 0.78 is not greater than 1.3277, reject the null hypothesis.D) Because the computed value of t = 0.78 is not greater than 1.3277, do not reject the null hypothesis.

Q: For the following hypothesis:With n = 20, = 71.2, s = 6.9, and = 0.1, state the calculated value of the test statistic t.A) 1.58B) 0.78C) 1.14D) 0.41

Q: For the following hypothesis:With n = 20, = 71.2, s = 6.9, and = 0.1, state the decision rule in terms of the critical value of the test statistic.A) This is a one-tailed test of the population mean with unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is greater than 1.3277. Otherwise, reject.B) This is a one-tailed test of the population mean with unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is greater than 2.1727. Otherwise, reject.C) This is a one-tailed test of the population mean with unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is greater than 1.3277. Otherwise, do not reject.D) This is a one-tailed test of the population mean with unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is greater than 2.1727. Otherwise, do not reject.

Q: For the following hypothesis test: With n = 15, s = 7.5, and = 62.2, state the conclusion. A) Because the computed value of t = 0.878 is not less than -2.1448 and not greater than 2.1448, do not reject the null hypothesis. B) Because the computed value of t = 1.312 is not less than -2.1448 and not greater than 2.1448, do not reject the null hypothesis. C) Because the computed value of t = 0.878 is not less than -2.1448 and not greater than 2.1448, reject the null hypothesis D) Because the computed value of t = 1.312 is not less than -2.1448 and not greater than 2.1448, reject the null hypothesis

Q: For the following hypothesis test: With n = 15, s = 7.5, and = 62.2, state the calculated value of the test statistic t. A) 1.014 B) 0.012 C) 0.878 D) 1.312

Q: For the following hypothesis test:With n = 15, s = 7.5, and = 62.2, state the decision rule in terms of the critical value of the test statisticA) This is a two-tailed test of the population mean with unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than -2.1448 or greater than 2.1448. Otherwise, do not reject.B) This is a two-tailed test of the population mean with unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is less than -2.1448 or greater than 2.1448. Otherwise, do not accept.C) This is a two-tailed test of the population mean with unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than -1.1828 or greater than 1.1828. Otherwise, do not reject.D) This is a two-tailed test of the population mean with unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than -1.1828 or greater than 1.1828. Otherwise, do not accept.

Q: For the following hypothesis test:With n = 80, = 9, and = 47.1, state the conclusion.A) Because the computed value of z = 3.012 is greater than 2.05, accept the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02B) Because the computed value of z = 3.012 is greater than 2.05, reject the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02C) Because the computed value of z = 2.087 is greater than 2.05, accept the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02D) Because the computed value of z = 2.087 is greater than 2.05, reject the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02

Q: For the following hypothesis test:With n = 80, = 9, and = 47.1, state the appropriate p-value.A) 0.0183B) 0.0314C) 0.0512D) 0.0218

Q: For the following hypothesis test:With n = 80, = 9, and = 47.1, state the calculated value of the test statistic z.A) 3.151B) -2.141C) 2.087D) -3.121

Q: For the following z-test statistic, compute the p-value assuming that the hypothesis test is a one-tailed test: z = -1.55. A) 0.0606 B) 0.1512 C) 0.0901 D) 0.0172

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: 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 upper limit for a 95 percent confidence interval estimate is approximately 12.106 ounces.

Q: Confidence intervals constructed with small samples tend to have greater margins of error than those constructed from larger samples, all else being constant.

Q: When calculating a confidence interval, the reason for using the t-distribution rather than the normal distribution for the critical value is that the population standard deviation is unknown.

Q: If we are interested in estimating the population mean based on a sample from a population for which we know neither the mean nor the standard deviation, the critical value will be a t value from the t-distribution.

Q: The standard deviation for the checking account balances is assumed known to be $357.50. Recently, a bank manager was interested in estimating the mean balance. To do this, she selected a random sample of 81 accounts and found a mean balance of $1,347.20. At a 95 percent confidence level, the lower limit for the confidence interval is $646.50.

Q: One way to reduce the margin of error in a confidence interval estimate is to lower the level of confidence.

Q: The margin of error is one-half the width of the confidence interval.

Q: In developing a confidence interval estimate, the margin of error is directly dependent on the value of the point estimate.

Q: In a recent report to the supply-chain manager in a major electronics company, the report writer stated that with 90 percent confidence, the manufacturing lead time for a critical part is between 3.34 hours and 4.14 hours. Based on this information, the margin of error for this estimate is .80 hours.

Q: In a recent report to the supply-chain manager in a major electronics company, the report writer stated that with 90 percent confidence, the manufacturing lead time for a critical part is between 3.34 hours and 4.14 hours. Based on this, the sample mean that generated the confidence interval was 3.60.

Q: To find a 98 percent confidence interval where the standard deviation is known, the correct z-value to use for the critical value is 2.33.

Q: When developing a confidence interval estimate, the confidence level is calculated based on the size of the sample and the population standard deviation.

Q: In a situation where we know the population standard deviation but wish to estimate the population mean using a 90 percent confidence interval, the critical value is z = 1.645.

Q: A random sample of 100 boxes of cereal had a sample mean weight of 396 grams. The standard deviation is known to be 5 grams. The upper end of the confidence interval for the mean is 405.8 grams.

Q: A report in a consumer magazine indicated that with 90 percent confidence, the mean number of hours that a particular brand light bulb lasts is between 900 and 1,100 hours. Based on this, the sample mean that produced this estimate is 1,000 hours.

Q: Recently, a marketing research company reported that based on a random sample of 300 households the mean number of trips to a major shopping mall per month per household is 4.12 trips. This value is referred to as a parameter and is subject to sampling error.

Q: A 95 percent confidence interval estimate indicates that there is a 95 percent chance that the true population value will fall within the range defined by the upper and lower limits.

Q: If a population is skewed, the point estimate will be pushed to the right or left of the middle of the confidence interval estimate.

Q: When using a 95 percent confidence interval for a mean, the area in the upper tail of the distribution that is outside the interval is 5 percent.

Q: The fact that a point estimate will likely be different from the corresponding population value is due to the fact that point estimates are subject to sampling error.

Q: A point estimate for the population mean will always fall within the confidence interval estimate.

Q: To construct a 99 percent confidence interval where Ïƒ is known, the correct critical value is 1.96.

Q: A statement in the newspaper attributed to the leader of a local union stated that the average hourly wage for union members in the region is $13.35. He indicated that this number came from a survey of union members. If an estimate was developed with 95 percent confidence, we can safely conclude that this value is within 95 percent of the true population mean hourly wage.

Q: A confidence interval will contain the true population value as long as the point estimate is within the lower to upper limits.

Q: A point estimate is equally likely to be higher or lower than the population mean if the sampling is done using a statistical sampling procedure.

Q: The product manager for a large retail store has recently stated that she estimates that the average purchase per visit for the store's customers is between $33.00 and $65.00. The $33.00 and the $65.00 are considered point estimates for the true population mean.

Q: The higher the level of confidence, the wider the confidence interval must be.

Q: Sampling error is the difference between a statistic computed from a sample and the corresponding parameter computed from the population.

Q: A human resources manager wishes to estimate the proportion of employees in her large company who have supplemental health insurance. What is the largest size sample she should select if she wants 95 percent confidence and a margin of error of 0.01?

Q: Under what circumstances would you wish to select a pilot sample?

Q: A financial analyst is interested in estimating the proportion of publicly traded companies on the New York Stock Exchange that have cash balances that are more than 10 percent of the total assets of the company. A random sample of n = 100 companies shows that 13 had cash balances of more than 10 percent of assets. Based on this information, develop and interpret a 90 percent confidence interval estimate for the population proportion.

Q: If a manager is interested in estimating the mean time customers spend shopping in a store on each visit to the store, she may want to develop a confidence interval estimate. Suppose, she has determined the required sample size and feels that she cannot afford one that large. What options are available?

Q: One of the major oil products companies conducted a study recently to estimate the mean gallons of gasoline purchased by customers per visit to a gasoline station. To do this, a random sample of customers was selected with the following data being recorded that show the gallons of gasoline purchased.8.722.49.513.318.92214.435.71924.95.715.78.922.515.9Based on these sample data, construct and interpret a 95 percent confidence interval estimate for the population mean.

Q: What are the disadvantages of using a small sample to estimate the population mean?

Q: In estimating a population mean, under what conditions would the t-distribution be used?

Q: In discussing a confidence interval estimate for a population mean, is it acceptable to provide an interpretation like the following: "There is a 95 percent chance that Î¼ lies in the range 20 to 40"?

Q: In a recent audit report, an accounting firm stated that the mean sale per customer for the client was estimated to be between $14.50 and $28.50. Further, this was based on a random sample of 100 customers and was computed using 95 percent confidence. Provide a correct interpretation of this confidence interval estimate.

Q: The sampling distribution for a proportion has a formula for that standard error that involves using p. Yet when a confidence interval is calculated for a proportion, the standard error formula uses the sample proportion. Why do they differ?

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