Questions and Answers of OpenIntro Statistics

9.23 A company selling licenses for new e-commerce computer software advertises that firms using this software obtain, on average during the first year, a yield of 10% on their initial investments. A
9.22 On the basis of a random sample the null hypothesis H0 : m = m0 is tested against the alternative H1 : m 7 m0 and the null hypothesis is not rejected at the 5% significance level.a. Does this
9.14 Test the hypotheses H0 : m … 100 H1 : m 7 100 using a random sample of n = 36, a probability of Type I error equal to 0.05, and the following sample statistics.a. x = 108; s = 20b. x = 104; s
9.10 A random sample of n = 25 is obtained from a population with variance s2, and the sample mean is computed to be x = 70. Consider the null hypothesis H0 : m = 80 versus the alternative hypothesis
9.9 A random sample is obtained from a population with a variance of s2 = 400, and the sample mean is computed to be xc = 70. Consider the null hypothesis H0 : m = 80 versus the alternative
9.8 A random sample of n = 25 is obtained from a population with variance s2, and the sample mean is computed.Test the null hypothesis H0 : m = 120 versus the alternative hypothesis H1 : m 7 120 with
9.7 A random sample is obtained from a population with variance s2 = 625, and the sample mean is computed.Test the null hypothesis H0 : m = 100 versus the alternative hypothesis H1 : m 7 100 with a =
9.6 Here we have the complement of Exercise 9.5. The 2000 presidential election in the United States was very close, and the decision came down to the results of the presidential voting in the state
9.3 John Stull, senior vice president of manufacturing, is seeking strong evidence to support his hope that new operating procedures have reduced the percentage of underfilled cereal packages from
9.6 Tests of the Variance of a Normal Distribution
9.5 Assessing the Power of a Test Tests of the Mean of a Normal Distribution: Population Variance Known Power of Population Proportion Tests (Large Samples)
9.4 Tests of the Population Proportion (Large Samples)
9.3 Tests of the Mean of a Normal Distribution:Population Variance Unknown
9.2 Tests of the Mean of a Normal Distribution:Population Variance Known p-Value Two-Sided Alternative Hypothesis
9.1 Concepts of Hypothesis Testing
8.31 Brämhults Juice, based in Sweden, produces chilled fruit and vegetable juices. A manager at Brämhults is considering purchasing a new machine to bottle 16-fluid-ounce (473-milliliter) bottles
8.30 A tuition center employs two statistics professors. The head of the tuition center wants to know if there is a significant difference in the average final exam scores between the students taught
8.28 To increase the fertility of his crops, a maize farmer adds two types of fertilizers, A and B, to two different areas of his farmland. He selects a random sample of 11 ears of maize planted
8.27 Independent random samples from two normally distributed populations give the following results:nx = 8 x = 77 sx = 6 ny = 13 y = 82 sy = 5a. If we assume that the unknown population variances
8.26 Independent random samples from two normally distributed populations give the following results:nx = 22 x = 2500 sx = 250 ny = 20 y = 2100 sy = 100 If we do not assume that the unknown
8.25 Independent random samples from two normally distributed populations give the following results:nx = 10 x = 250 sx = 14 ny = 17 y = 165 sy = 23 Assume that the unknown population variances are
8.19 Calculate the 95% confidence interval for the difference in population proportions for each of the following:a. nx = 260 pn x = 0.35 ny = 240 pn y = 0.30b. nx = 145 pn x = 0.20 ny = 120 pn y =
8.18 Calculate the margin of error for each of the following:a. nx = 300 pn x = 0.62 ny = 350 pn y = 0.72b. nx = 100 pn x = 0.44 ny = 120 pn y = 0.55
8.11 Determine the margin of error for a 95% confidence interval for the difference between population means for each of the following (assume equal population variances):a. nx = 100 s2x= 36 x = 300
8.10 Assuming unequal population variances, determine the number of degrees of freedom for each of the following:a. nx = 16 s2x= 5 ny = 4 s2y= 36b. nx = 9 s2x= 30 ny = 16 s2y= 4
8.8 Assuming equal population variances, determine the number of degrees of freedom for each of the following:a. nx = 16 s2x= 30 ny = 9 s2y= 36b. nx = 12 s2x= 30 ny = 14 s2y= 36c. nx = 20 s2x= 16 ny
8.7 Independent random sampling from two normally distributed populations gives the following results:nx = 81; x = 140; s2x= 25 ny = 100; y = 120; s2y= 14 Find a 95% confidence interval estimate of
8.6 Independent random sampling from two normally distributed populations gives the following results:nx = 87; x = 115; sx = 25 ny = 88; y = 109; sy = 15 Find a 99% confidence interval estimate of
8.5 A random sample of six salespeople who attended a motivational course on sales techniques was monitored 3 months before and 3 months after the course.The table shows the values of sales (in
8.4 A random sample of 10 pairs of identical houses was chosen in a large Midwestern city, and a passive solar heating system was installed in one house from each pair. The total fuel bills (in
8.3 An educational study was designed to investigate the effectiveness of a reading program of elementary age children. Each child was given a pretest and posttest.Higher posttest scores would
8.1 A dependent random sample from two normally distributed populations gives the following results:n = 11 d = 28.5 sd = 3.3a. Find the 98% confidence interval for the difference between the means of
8.3 Confidence Interval Estimation of the Difference Between Two Population Proportions (Large Samples)
8.2 Confidence Interval Estimation of the Difference Between Two Normal Population Means: Independent Samples Two Means, Independent Samples, and Known Population Variances Two Means, Independent
8.1 Confidence Interval Estimation of the Difference Between Two Normal Population Means: Dependent Samples
7.94 Japan’s rainy season starts in July and continues through August. From a random sample of 40 days in 2021, the mean temperature was around 26.3°C with a standard deviation of 1.6°C. Compute
7.92 Consider the data in Exercise 7.90. It was reported in the local paper that almost two-thirds (from 57%to 63%) of the population prefers the online renewal process. What is the confidence level
7.91 Consider the data in Exercise 7.90. Suppose that we computed for the population proportion who pay for vehicle registration by mail a confidence interval extending from 0.15 to 0.17. What is the
7.89 The following data represent the scores obtained from a random sample of 20 people who attempted an IQ test having a maximum score of 100:85 92 54 46 98 70 63 56 95 40 57 98 81 47 72 40 68 41 70
7.88 A restaurant conducted a survey to determine whether customers are satisfied with its food delivery service by clicking “Yes” or “No” on an online form shared with them. From a random
7.85 A teacher selected a random sample of 170 primary students from her school. They were asked what amount they received as daily pocket money. The sample mean response was €2.50, and the sample
7.83 Students’ scores on the final statistics test from the previous semester at a university were taken for a random sample of 9 students and were recorded as follows:69 92 82 30 73 91 48 74 62a.
7.82 Suppose the manager of Croquettenbar Smaeck, a restaurant in Antwerp, Belgium, is estimating the queuing time of the customers until they receive their meals during peak hours. From a random
7.81 The following data represent the exchange rate of Malaysian ringgit (RMY) to Hong Kong dollar (HKD), retrieved from the Interbank Foreign Exchange Market in Kuala Lumpur, Malaysia, from
7.79 Students at a university in Ghent, Belgium, are expected to complete a group presentation in 20 minutes as one of the assessments to pass the business module. In a random sample of 18
7.78 Wosiwosi, a UK-based African food store, specializes in the distribution of wholesale and retail of African and Caribbean food. Suppose the manager wants to estimate the mean amount customers
7.73 Determine the sample size needed for each of the following situations.a. N = 5,000; pn = 0.5; 1.96spn = 0.10b. N = 5,000; pn = 0.5; 1.96spn = 0.05 Application Exercises
7.72 Determine the sample size needed for each of the following situations.a. N = 1,000; s = 450; 1.96sx = 50b. N = 3,000; s = 450; 1.96sx = 50c. N = 1,000,000; s = 450; 1.96sx = 50d. Compare and
7.71 Determine the sample size needed for each of the following situations.a. N = 785; s = 24; 1.96sx = 5b. N = 785; s = 24; 1.96sx = 88c. N = 785; s = 24; 1.96sx = 190d. Compare and comment on your
7.67 How large a sample is needed to estimate the population proportion for each of the following?a. ME = 0.05; a = 0.01b. ME = 0.05; a = 0.10c. Compare and comment on your answers to parts a and
7.66 How large a sample is needed to estimate the population proportion for each of the following?a. ME = 0.03; a = 0.05b. ME = 0.05; a = 0.05c. Compare and comment on your answers to parts a and b.
7.55 Take a random sample of 80 pages from this book and estimate the proportion of all pages that contain mathematical formulas.
7.54 Assume simple random sampling. Calculate the confidence interval for the population proportion, P, for each of the following.a. N = 1,675; n = 400; x = 120; 99% confidence levelb. N = 955; n =
7.51 Assume simple random sampling. Calculate the variance of the sample mean, s2x, for each of the following.a. N = 1,300; n = 100; s = 15b. N = 1,500; n = 200; s2 = 91c. N = 3,000; n = 220; s2 = 150
7.50 A watch manufacturer in Switzerland uses sputtered titanium coating for some models it creates. A random sample of nine observations on the thickness of this coating is taken from a week’s
7.42.7.44 Consider the following random sample from a normal population:11 17 8 7 9a. Find the 90% confidence interval for population variance.b. Find the 95% confidence interval for the population
7.42 Find the lower confidence limit for the population variance for each of the following normal populations.a. n = 21; a = 0.05; s2 = 16b. n = 16; a = 0.05; s = 8c. n = 28; a = 0.01; s = 15 7.43
7.32 A small private university is planning to start a volunteer football program. A random sample of alumni is surveyed. It was found that 250 were in favor of this program, 75 were opposed, and 25
7.31 Calculate the confidence interval to estimate the population proportion for each of the following.a. 90% confidence level; n = 700; pn = 0.20b. 99% confidence level; n = 140; pn = 0.01c. a =
7.30 Find the margin of error to estimate the population proportion for each of the following.a. n = 350; pn = 0.30; a = 0.01b. n = 275; pn = 0.45; a = 0.05c. n = 500; pn = 0.05; a = 0.10
7.29 Build Rental is a plant hire company in the United Kingdom that provides machinery, equipment, and tools for a limited period to construction contractors.It is interested in the amount of time
7.27 Weight Medics, a clinic in the United Kingdom, offers a weight-loss program for its clients. A review of its records found the following amounts of weight loss, in pounds, for a random sample of
7.26 There is concern about the speed of automobiles traveling over a particular stretch of highway. For a random sample of 28 automobiles, radar indicated the following speeds, in miles per hour:59
7.23 The Programme for International Student Assessment(PISA) is a global study conducted by the Organization for Economic Co-operation and Development(OECD). It measures mathematical performance as
7.22 Calculate the width for each of the following.a. n = 6; s = 40; a = 0.05b. n = 22; s2 = 400; a = 0.01c. n = 25; s = 50; a = 0.10 Application Exercises
7.20 Find the LCL and UCL for each of the following.a. a = 0.05; n = 25; x = 560; s = 45b. a>2 = 0.05; n = 9; x = 160; s2 = 36c. 1 - a = 0.98; n = 22; x = 58; s = 15
7.17 Find the standard error to estimate the population mean for each of the following.a. n = 17; 95% confidence level; s = 16b. n = 25; 90% confidence level; s2 = 43 7.18 Calculate the margin of
7.12 Assume a normal distribution with known population variance. Calculate the LCL and UCL for each of the following.a. x = 255; n = 304; s = 30; a = 0.01b. x = 395; n = 105; s2 = 100; a = 0.05c. x
7.11 Assume a normal distribution with known population variance. Calculate the width to estimate the population mean, m, for the following.a. 90% confidence level; n = 100; s2 = 169b. 95% confidence
7.10 Assume a normal distribution with known population variance. Calculate the margin of error to estimate the population mean for the following.a. 99% confidence level; n = 81; s2 = 169b. 90%
7.9 Find the reliability factor, za 2, to estimate the mean, m, of a normally distributed population with known population variance for the following.a. a = 0.09b. a2 = 0.01
7.7 Suppose that x1 and x2 are random samples of observations from a population with mean m and variance s2. Consider the following three point estimators, X, Y, Z, of m:X =1 2x1 +1 2x2 Y =1 4x1 +3
7.6 According to CleanTechnica website, Sweden’s full electric vehicles took a record high 24.1%share of new sales, with the Kia e-Niro being Sweden’s best-selling full electric vehicle. Suppose
7.4 A random sample of 15 employees in a large manufacturing plant found the following figures for number of hours of overtime worked in the last month:24 16 17 18 21 17 16 19 19 24 15 19 18 21 24
7.3 A random sample of 10 economists produced the following forecasts for percentage unemployment rate in the next year:4.4 4.2 3.7 3.9 4.4 3.9 3.7 4.0 4.1 4.0a. The population meanb. The population
7.2 A random sample of 10 homes in a particular suburb had the following selling prices (in thousands of pounds):209 256 257 272 275 277 278 285 306 327a. Check for evidence of nonnormality.b. Find a
7.1 There is concern about the speed of automobiles traveling over a particular stretch of highway. For a random sample of 28 automobiles, radar indicated the following speeds, in miles per hour:58
7.8 Sample-Size Determination: Finite Populations Sample Sizes for Simple Random Sampling: Estimation of the Population Mean or Total Sample Sizes for Simple Random Sampling: Estimation of Population
7.7 Sample-Size Determination: Large Populations Mean of a Normally Distributed Population, Known Population Variance Population Proportion
7.6 Confidence Interval Estimation: Finite Populations Population Mean and Population Total Population Proportion
7.5 Confidence Interval Estimation for the Variance of a Normal Distribution
7.4 Confidence Interval Estimation for Population Proportion(Large Samples)
7.3 Confidence Interval Estimation for the Mean of a Normal Distribution:Population Variance Unknown Student’s t Distribution Intervals Based on the Student’s t Distribution
7.2 Confidence Interval Estimation for the Mean of a Normal Distribution:Population Variance Known Intervals Based on the Normal Distribution Reducing Margin of Error
7.1 Properties of Point Estimators Unbiased Most Efficient
6.76 Scores on a particular test, taken by a large group of students, follow a normal distribution with a variance of 2500. A random sample of 25 scores was taken to estimate the population’s mean
6.73 The length of the Atlantic salmon sold by a fishmonger has a normal distribution with a standard deviation of 2 inches. A random sample of 9 Atlantic salmons is selected from the fishmonger.a.
6.72 It is found that out of a group of elders, 65% have high glucose levels. Suppose their glucose level is normally distributed with a mean of 13 mmol/L and a standard deviation of 1.5 mmol/L.a.
6.71 It has been found the retirement age of a community follows a normal distribution with a mean of 75 years and a standard deviation of 3 years. A random sample of 11 people from the community is
6.70 Refer to the chapter appendix in order to derive the mean of the sampling distribution of the sample variances for a sample of n observations from a population of N members when the population
6.69 The gross area values of the lands sold by a property agency are known to be normally distributed with a mean of 999 acres and a standard deviation of 250 acres.a. For a random sample of 8
6.68 A bank web operation officer is monitoring online credit card transactions and determines that the transaction duration follows a normal distribution with a mean of 20 seconds and a standard
6.67 A flour manufacturer found that the weightage error of bags labeled 2 kilograms of flour follows a normal distribution with a mean of 10 grams and a standard deviation of 2 grams. A random
6.66 A study determines that the height of Taiwanese men aged 65 or more is normally distributed with an average of 163.8 centimeters and standard deviation of 6.2 centimeters. A random sample of 20
6.65 Explain how the central limit theorem works on the sampling distribution of the sample means from a population that is normally distributed and one that is not normally distributed.
6.63 What is meant by the statement that the sample mean and the sample proportion have a sampling distribution?What other statistical examples can you provide?
6.52 It is believed that first-year salaries for newly qualified accountants follow a normal distribution with a standard deviation of $2,500. A random sample of 16 observations was taken.a. Find the
6.50 The athletic association of a school reports that the men’s 100-meter run follows a normal distribution with a variance of 1.58 seconds. A random sample of 23 records of the men’s 100-meter
6.49 A random sample of size n = 11 is obtained from a normally distributed population with a population mean of m = 27 and a variance of s2 = 16.a. What is the probability that the sample mean is
6.48 A random sample of size n = 16 is obtained from a normally distributed population with a population mean of m = 114 and a variance of s2 = 484.a. What is the probability that the sample mean is

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