MATH 225N WEEK 6 QUIZ STATISTICS – QUESTION AND ANSWERS

MATH 225N WEEK 6 QUIZ STATISTICS – QUESTION AND ANSWERS Question 1 A statistics professor recently graded final exams for students in her introductory statistics course. In a review of her grading, she found the mean score out of 100 points was a x¯=77, with a margin of error of 10. Construct a confidence interval for the mean score (out of 100 points) on the final exam. ________________________________________ That is correct! ________________________________________ $$(67, 87) Answer Explanation Correct answers: • $left(67, 87right)$(67, 87) A confidence interval is an interval of values, centered on a point estimate, of the form (pointestimate−marginof error,pointestimate+marginof error) Using the given point estimate for the mean, x¯=77 and margin of error 10, the confidence interval is: (77−10,77+10)(67,87) FEEDBACK • • • • Content attribution- Opens a dialog Question 2 A random sample of adults were asked whether they prefer reading an e-book over a printed book. The survey resulted in a sample proportion of p′=0.14, with a sampling standard deviation of σp′=0.02, who preferred reading an e-book. Use the empirical rule to construct a 95% confidence interval for the true proportion of adults who prefer e-books. ________________________________________ That is correct! ________________________________________ $$(0.1, 0.18) Answer Explanation Correct answers: • $left(0.10, 0.18right)$(0.10, 0.18) By the Empirical Rule, a 95% confidence interval corresponds to a z-score of z=2. Substituting the given values p′=0.14and σp′=0.02, a confidence interval is (p′−z⋅σp′,p′+z⋅σp′)(0.14−2⋅0.02,0.14+2⋅0.02)(0.14−0.04,0.14+0.04)(0.10,0.18) FEEDBACK • • • • Content attribution- Opens a dialog Question 3 The pages per book in a library are normally distributed with an unknown population mean. A random sample of books is taken and results in a 95% confidence interval of (237,293) pages. What is the correct interpretation of the 95% confidence interval? ________________________________________ That is correct! ________________________________________ We estimate with 95% confidence that the sample mean is between 237 and 293 pages. We estimate that 95% of the time a book is selected, there will be between 237 and 293 pages. We estimate with 95% confidence that the true population mean is between 237 and 293 pages. Answer Explanation Correct answer: We estimate with 95% confidence that the true population mean is between 237 and 293 pages. Once a confidence interval is calculated, the interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated, and state the confidence interval. We estimate with 95% confidence that the true population mean is between 237 and 293 pages. FEEDBACK • • • • Content attribution- Opens a dialog Question 4 The population standard deviation for the heights of dogs, in inches, in a city is 3.7 inches. If we want to be 95% confident that the sample mean is within 2 inches of the true population mean, what is the minimum sample size that can be taken? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table above for the z-score, and be sure to round up to the nearest integer. ________________________________________ That is correct! ________________________________________ $$14 dog heights Answer Explanation Correct answers: • $14text{ dog heights}$14 dog heights The formula for sample size is n=z2σ2EBM2. In this formula, z=zα2=z0.025=1.96, because the confidence level is 95%. From the problem, we know that σ=3.7 and EBM=2. Therefore, n=z2σ2EBM2=(1.96)2(3.7)222≈13.15. Use n=14 to ensure that the sample size is large enough. Also, the sample size formula shown above is sometimes written using an alternate format of n=(zσE)2. In this formula, E is used to denote margin of error and the entire parentheses is raised to the exponent 2. Therefore, the margin of error for the mean can be denoted by "EBM" or by "E". Either formula for the sample size can be used and these formulas are considered as equivalent. FEEDBACK • • • • Content attribution- Opens a dialog Question 5 Clarence wants to estimate the percentage of students who live more than three miles from the school. He wants to create a 98% confidence interval which has an error bound of at most 4%. How many students should be polled to create the confidence interval? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576Use the table of values above. ________________________________________ That's not right. ________________________________________ $$847 students Answer Explanation Correct answers: • $846 students$846 students Given the information in the question, EBP=0.04 since 4%=0.04 and zα2=z0.01=2.326 because the confidence level is 98%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore, n=z2p′q′EBP2=2.3262(0.5)(0.5)0.042=845.4 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 846students. FEEDBACK • • • • Content attribution- Opens a dialog Question 6 The average score of a random sample of 87 senior business majors at a university who took a certain standardized test follows a normal distribution with a standard deviation of 28. Use Excel to determine a 90% confidence interval for the mean of the population. Round your answers to two decimal places and use ascending order. Score 516 536 462 461 519 496 517 488 521 HelpCopy to ClipboardDownload CSV ________________________________________ That is correct! ________________________________________ $$(509.30, 519.18) Answer Explanation Correct answers: • $left(509.30, 519.18right)$(509.30, 519.18) A 90% confidence interval for μ is (x¯−zα/2σn‾√,x¯+zα/2σn‾√). Here, α=0.1, σ=28, and n=87. Use Excel to calculate the 90% confidence interval. 1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean, rounded to two decimal places, is x¯=514.24. 2. Click on any empty cell, enter =CONFIDENCE.NORM(0.1,28,87), and press ENTER. 3. The margin of error, rounded to two decimal places, is zα/2σn‾√≈4.94. The confidence interval for the population mean has a lower limit of 514.24−4.94=509.30 and an upper limit of 514.24+4.94=519.18. Thus, the 90% confidence interval for μ is (509.30, 519.18). FEEDBACK • • • • Content attribution- Opens a dialog Question 7 A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each one recorded. The results are given below. Assume the percentages of students' absences are approximately normally distributed. Use Excel to estimate the mean percentage of absences per tutorial over the past 5 years with 90% confidence. Round your answers to two decimal places and use increasing order. Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 HelpCopy to ClipboardDownload CSV ________________________________________ That's not right. ________________________________________ $$(9.22, 11.610.) Answer Explanation Correct answers: • $left(9.22, 11.60right)$(9.22, 11.60) Use Excel to calculate the 90% confidence interval, where α=0.1 and n=28. 1. Open Excel and enter the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, using the STDEV.S function. Thus, the sample mean is 10.41 and the sample standard deviation, rounded to two decimal places, is 3.71. 2. Click on any empty cell, enter =CONFIDENCE.T(0.1,3.71,28), and press ENTER. 3. The margin of error, rounded to two decimal places, is 1.19. The confidence interval for the population mean has a lower limit of 10.41−1.19=9.22 and an upper limit of 10.41+1.19=11.60. Thus, the 90% confidence interval for the mean percentage of absences per tutorial over the past 5 years is (9.22, 11.60). FEEDBACK • • • • Content attribution- Opens a dialog Question 8 Eric is studying people's typing habits. He surveyed 525 people and asked whether they leave one space or two spaces after a period when typing. 440 people responded that they leave one space. Create a 90% confidence interval for the proportion of people who leave one space after a period. • Round your results to four decimal places. ________________________________________ That is correct! ________________________________________ ($$.8117, $$.8645) Answer Explanation (1$$, 2$$) Correct answers: • 1$0.8117$0.8117 • 2$0.8645$0.8645 The confidence interval for the unknown population proportion p is (p̂ −z⋆p̂ (1−p̂ )n‾‾‾‾‾‾‾‾‾√,p̂ −z⋆p̂ (1−p̂ )n‾‾‾‾‾‾‾‾‾√). The confidence interval can be calculated using Excel. 1. Identify α. Click on cell A1 and enter =1−0.90 and press ENTER. 2. Thus, α=0.1. Enter the number of successes, x=440, and sample size, n=525, in the Excel sheet in cells A2 and A3. To find the proportion of successes, p̂ , click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p̂ ≈0.8381. Use the NORM.S.INV function in Excel to find z⋆. Click on cell A5, enter =NORM.S.INV(1−A1/2), and press ENTER. 4. The answer for z⋆, rounded to two decimal places, is z⋆≈1.64. To calculate the standard error, p̂ (1−p̂ )n‾‾‾‾‾‾‾‾‾√, click on cell A6 and enter =SQRT(A4∗(1−A4)/A3) and press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p̂ (1−p̂ )n‾‾‾‾‾‾‾‾‾√≈0.0161. To calculate the margin of error, z⋆p̂ (1−p̂ )n‾‾‾‾‾‾‾‾‾√, click on cell A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is z⋆p̂ (1−p̂ )n‾‾‾‾‾‾‾‾‾√≈0.0264. The confidence interval for the population proportion has a lower limit of A4−A7=0.8117 and an upper limit of A4+A7=0.8645. Thus, the 90%confidence interval for the population proportion of people who leave one space after a period, based on this sample, is approximately (0.8117, 0.8645). FEEDBACK • • • • Content attribution- Opens a dialog Question 9 A sample of 27 employees for the Department of Health and Human Services has the following salaries, in thousands of dollars. Assuming normality, use Excel to find the 98% confidence interval for the true mean salary, in thousands of dollars. Round your answers to two decimal places and use increasing order. Salary 71 70 69 65 72 69 72 72 71 HelpCopy to ClipboardDownload CSV ________________________________________ That is correct! ________________________________________ $$(69.14, 71.38) Answer Explanation Correct answers: • $left(69.14, 71.38right)$(69.14, 71.38) Use Excel to calculate the 98% confidence interval, where α=0.02 and n=27. 1. Open Excel and enter the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, using the STDEV.S function. Thus, the sample mean, rounded to two decimal places, is 70.26 and the sample standard deviation, rounded to two decimal places, is 2.35. 2. Click on any empty cell, enter =CONFIDENCE.T(0.02,2.35,27), and press ENTER. 3. The margin of error, rounded to two decimal places, is 1.12. The confidence interval for the population mean has a lower limit of 70.26−1.12=69.14 and an upper limit of 70.26+1.12=71.38. Thus, the 98% confidence interval for the true mean salary, in thousands of dollars, is (69.14, 71.38). FEEDBACK • • • • Content attribution- Opens a dialog Content attribution- Opens a dialog Question 11 A random sample of house sizes in major city has a sample mean of x¯=1204.9 sq ft and sample standard deviation of s=124.6 sq ft. Use the Empirical Rule to determine the approximate percentage of house sizes that lie between 955.7and 1454.1 sq ft. Round your answer to the nearest whole number (percent). ________________________________________ That is correct! ________________________________________ $$95 $%$% Answer Explanation 1$$ $%$% Correct answers: • 1$95$95 To use the Empirical Rule, we need to know how many standard deviations from the mean are the given values 955.7 and 1454.1. Since the mean is 1204.9, we see that the value 955.7 is 1204.9−955.7=249.2 sq ft below the mean. This is 2 standard deviations, since 249.2=2×124.6, so 955.7 is 2 standard deviations less than the mean. Similarly, the value 1454.1 is 1454.1−1204.9=249.2 sq ft above the mean. Again, this is 2 standard deviations, since 249.2=2×124.6, so 1454.1 is 2 standard deviations greater than the mean. The Empirical Rule states that approximately 95% of the data is within 2 standard deviations of the mean. So by the Empirical Rule, we can say that approximately 95% of house sizes in the major city are between 955.7 and 1454.1 sq ft. FEEDBACK • • • • Content attribution- Opens a dialog Question 12 The graph below shows the graphs of several normal distributions, labeled A, B, and C, on the same axis. Determine which normal distribution has the smallest standard deviation. A figure consists of three curves along a horizontal axis, labeled Upper A, Upper B and Upper C. Curve Upper A is short and the most spread out, curve Upper B is tall and the least spread out, and curve C is farther to the left than A. ________________________________________ That is correct! ________________________________________ A B C Answer Explanation Correct answer: B Remember that the standard deviation tells how spread out the normal distribution is. So a high standard deviation means the graph will be short and spread out. A low standard deviation means the graph will be tall and skinny. The distribution that is the tallest and least spread out is B, so that has the smallest standard deviation. FEEDBACK • • • • Content attribution- Opens a dialog Question 13 The resistance of a strain gauge is normally distributed with a mean of 100 ohms and a standard deviation of 0.3 ohms. To meet the specification, the resistance must be within the range 100±0.7 ohms. What proportion of gauges is acceptable? • Round your answer to four decimal places. ________________________________________ That is correct! ________________________________________ $$.9804 Answer Explanation Correct answers: • $0.9804$0.9804 The mean is μ=100, and the standard deviation is σ=0.3. As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 100.7 for X, 100 for Mean, 0.3 for Standard_dev, and TRUE for Cumulative, all for the higher value of X. Thus, the answer, rounded to four decimal places, is 0.9902. 5. Click on any other empty cell. Click Insert function, fx. 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 99.3 for X, 28 for Mean, 0.3 for Standard_dev, and TRUE for Cumulative, all for the lower value of X. Thus, the answer, rounded to four decimal places, is 0.0098. Now subtract, 0.9902−0.0098=0.9804. Thus, the probability that a random gauge is acceptable is 0.9804. FEEDBACK • • • • Content attribution- Opens a dialog Question 14 A baker knows that the daily demand for strawberry pies is a random variable that follows the normal distribution with a mean of 31.8 pies and a standard deviation of 4.5 pies. Find the demand that has an 8% probability of being exceeded. • Use Excel, and round your answer to two decimal places. ________________________________________ That is correct! ________________________________________ $$38.12 Answer Explanation Correct answers: • $38.12$38.12 Here, the mean, μ, is 31.8 and the standard deviation, σ, is 4.5. Let x be the number of pies such that there is an 8%probability that the demand will be above this number. The area to the right of x is 8%=0.08. So, the area to the left of xis 1−0.08=0.92. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.92,31.8,4.5) and press ENTER. The answer, rounded to two decimal places, is x≈38.12. Therefore, there is an 8% probability that the demand exceeds 38.12 pies. FEEDBACK • • • • Content attribution- Opens a dialog Question 15 A group of friends has gotten very competitive with their board game nights. They have found that overall, they each have won an average of 18 games, with a population standard deviation of 6 games. If a sample of only 2 friends is selected at random from the group, select the expected mean and the standard deviation of the sampling distribution from the options below. Remember to round to the nearest whole number. ________________________________________ That is correct! ________________________________________ • σx¯=6 games • ________________________________________ • σx¯=3 games • ________________________________________ • σx¯=4 games • ________________________________________ • μx¯=18 games • ________________________________________ • μx¯=3 games • ________________________________________ • μx¯=9 games • ________________________________________ Answer Explanation Correct answer: σx¯=4 games μx¯=18 games The standard deviation of the sampling distribution σx¯=σn√=62√≈4 games. When the distribution is normal, the mean of the sampling distribution is equal to the mean of the population μx¯=μ=18games. FEEDBACK • • • • Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Question 21 In order to estimate the average electricity usage per month, a sample of 125 residential customers were selected, and the monthly electricity usage was determined using the customers' meter readings. Assume a population variance of 12,100kWh2. Use Excel to find the 98% confidence interval for the mean electricity usage in kilowatt hours. Round your answers to two decimal places and use ascending order. Electric Usage 765 1139 714 687 1027 1109 749 799 911 HelpCopy to ClipboardDownload CSV ________________________________________ That is correct! ________________________________________ $$(894.43, 940.21) Answer Explanation Correct answers: • $left(894.43, 940.21right)$(894.43, 940.21) A 98% confidence interval for μ is (x¯−zα/2σn‾√,x¯+zα/2σn‾√). Here, α=0.02, σ=110, and n=125. Use Excel to calculate the 98% confidence interval. 1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean is x¯=917.32. 2. Click on any empty cell, enter =CONFIDENCE.NORM(0.02,110,125), and press ENTER. 3. The margin of error, rounded to two decimal places, is zα/2σn‾√≈22.89. The confidence interval for the population mean has a lower limit of 917.32−22.89=894.43 and an upper limit of 917.32+22.89=940.21. Thus, the 98% confidence interval for μ is (894.43, 940.21). FEEDBACK • • • • Content attribution- Opens a dialog Question 22 Hugo averages 72 words per minute on a typing test with a standard deviation of 13 words per minute. Suppose Hugo's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then, X∼N(72,13). Suppose Hugo types 69 words per minute in a typing test on Wednesday. The z-score when x=69 is ________. This z-score tells you that x=69 is ________ standard deviations to the ________ (right/left) of the mean, ________. Correctly fill in the blanks in the statement above. ________________________________________ That's not right. ________________________________________ Suppose Hugo types 69 words per minute in a typing test on Wednesday. The z-score when x=69 is −0.231. This z-score tells you that x=69 is 0.231 standard deviations to the left of the mean, 72. Suppose Hugo types 69 words per minute in a typing test on Wednesday. The z-score when x=69 is 0.188. This z-score tells you that x=69 is 0.188 standard deviations to the right of the mean, 72. Suppose Hugo types 69 words per minute in a typing test on Wednesday. The z-score when x=69 is 0.231. This z-score tells you that x=69 is 0.231 standard deviations to the right of the mean, 72. Suppose Hugo types 69 words per minute in a typing test on Wednesday. The z-score when x=69 is −0.188. This z-score tells you that x=69 is 0.188 standard deviations to the left of the mean, 72. Answer Explanation Correct answer: Suppose Hugo types 69 words per minute in a typing test on Wednesday. The z-score when x=69 is −0.231. This z-score tells you that x=69 is 0.231 standard deviations to the left of the mean, 72. The z-score can be found using the formula z=x−μσ=69−7213=−313≈−0.231 A negative value of z means that that the value is below (or to the left of) the mean, which was given in the problem as μ=72 words per minute in a typing test. The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. So, typing 69 words per minute is 0.231 standard deviations away from the mean. Your answer: Suppose Hugo types 69 words per minute in a typing test on Wednesday. The z-score when x=69 is 0.231. This z-score tells you that x=69 is 0.231 standard deviations to the right of the mean, 72. FEEDBACK • • • • Content attribution- Opens a dialog Question 23 Hugo averages 50 words per minute on a typing test with a standard deviation of 6 words per minute. Suppose Hugo's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then, X∼N(50,6). Suppose Hugo types 58 words per minute in a typing test on Wednesday. The z-score when x=58 is ________. This z-score tells you that x=58 is ________ standard deviations to the ________ (right/left) of the mean, ________. Correctly fill in the blanks in the statement above. ________________________________________ That's not right. ________________________________________ Suppose Hugo types 58 words per minute in a typing test on Wednesday. The z-score when x=58 is −0.889. This z-score tells you that x=58 is 0.889 standard deviations to the left of the mean, 50. Suppose Hugo types 58 words per minute in a typing test on Wednesday. The z-score when x=58 is −1.333. This z-score tells you that x=58 is 1.333 standard deviations to the left of the mean, 50. Suppose Hugo types 58 words per minute in a typing test on Wednesday. The z-score when x=58 is 0.889. This z-score tells you that x=58 is 0.889 standard deviations to the right of the mean, 50. Suppose Hugo types 58 words per minute in a typing test on Wednesday. The z-score when x=58 is 1.333. This z-score tells you that x=58 is 1.333 standard deviations to the right of the mean, 50. Answer Explanation Correct answer: Suppose Hugo types 58 words per minute in a typing test on Wednesday. The z-score when x=58 is 1.333. This z-score tells you that x=58 is 1.333 standard deviations to the right of the mean, 50. The zz-score can be found using the formula begin{align} z &= dfrac{x - mu}{sigma} & = dfrac{58 - 50}{6} & = dfrac{8}{6} & approx 1.333 end{align}A positive value of zz means that that the value is above (or to the right of) the mean, which was given in the problem as mu = 50μ=50 words per minute in a typing test. The zz-score tells you how many standard deviations the value xx is above (to the right of) or below (to the left of) the mean, muμ. So, typing 58 words per minute is 1.333 standard deviations away from the mean. Your answer: Suppose Hugo types $_58$_ words per minute in a typing test on Wednesday. The $_z$_-score when $_x = 58$_ is $_mathbf{0.889}$_. This $_z$_-score tells you that $_x = 58$_ is $_mathbf{0.889}$_ standard deviations to the right of the mean, $_mathbf{50}$_. FEEDBACK • • • • Content attribution- Opens a dialog Question 24 Lisa has collected data to find that the number of pages per book on a book shelf has a normal distribution. What is the probability that a randomly selected book has fewer than $_170$_ pages if the mean is $_190$_ pages and the standard deviation is $_20$_ pages? Use the empirical rule. Enter your answer as a percent rounded to two decimal places if necessary. ________________________________________ That is correct! ________________________________________ $$16% Answer Explanation Correct answers: • $16%$16% Notice that $_170$_ pages is one standard deviation less than the mean. Based on the empirical rule, $_68%$_ of the number of pages in the books are within one standard deviation of the mean. Since the normal distribution is symmetric, this implies that $_16%$_ of the number of pages in the books are less than one standard deviation less than the mean. FEEDBACK • • • • Content attribution- Opens a dialog Question 25 Lisa has collected data to find that the number of pages per book on a book shelf has a normal distribution. What is the probability that a randomly selected book has fewer than $_142$_ pages if the mean is $_190$_ pages and the standard deviation is $_24$_ pages? Use the empirical rule.Enter your answer as a percent rounded to two decimal places if necessary. ________________________________________ That is correct! ________________________________________ $$2.5% Answer Explanation Correct answers: • $2.5%$2.5% Notice that $_142$_ pages is two standard deviations less than the mean. Based on the empirical rule, $_95%$_ of the number of pages in the books are within two standard deviations of the mean. Since the normal distribution is symmetric, this implies that $_2.5%$_ of the number of pages in the books are less than two standard deviations less than the mean. FEEDBACK • • • • Content attribution- Opens a dialog