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The construction of confidence intervals for the mean score on a final exam and the true proportion of adults who prefer reading e-books over printed books. It also covers the interpretation of confidence intervals, the use of the empirical rule, and the calculation of sample size for a desired margin of error. Additionally, the document explores the application of normal distribution to determine the probability of a randomly selected book having fewer pages than a given threshold. The content covers key statistical concepts and techniques that are commonly encountered in various academic disciplines, particularly in fields involving data analysis, research methodology, and quantitative decision-making.

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Download Confidence Intervals and Normal Distribution in Statistics and more Exams Nursing in PDF only on Docsity! 1 Question 1 1/1 points 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 100p o i n t s was a x̄=77, with a margin of error of 10. Construct a confidence interval for the mean score (out of 100p o i n t s ) on the final exam. That is correct! open paren 67 comma 87 close paren$$ open paren 67 comma 87 close paren - correct Answer Explanation Correct answers: • open paren 67 comma 87 close paren $\left(67,\ 87\right)$ • A confidence interval is an interval of values, centered on a point estimate, of the form (pointestimate−marginoferror,pointestimate+marginoferror) Using the given point estimate for the mean, x̄=77and margin of error 10, the confidence interval is: 2 (77−10,77+10)(67,87) • • • • 5 • Question 3 · 1/1 points The pages per book in a library are normally distributed with an unknown population mean. A random sample of books is taken and results in a 95%confidence interval of (237,293)pages. What is the correct interpretation of the 95%conf idence interval? That is correct! We estimate with 95%conf i dence that the sample mean is between 237and 293p a g e s . We estimate that 95%of the time a book is selected, there will be between 237and 293p a g e s . We estimate with 95%confidence that the true population mean is between 237and 293 pages. 6 Answer Explanation Correct answer: We estimate with 95%confidence that the true population mean is between 237and 293 pages. 7 Once a confidence interval is calculated, the interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated, and state the confidence interval. We estimate with 95%confidence that the true population mean is between 237and 293 pages. • • • • Question 4 · 1/1 points The population standard deviation for the heights of dogs, in inches, in a city is 3.7i n c h e s . If we want to be 95%confident that the sample mean is within 2inches of the true population mean, what is the minimum sample size that can be taken? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table above for the z-score, and be sure to round up to the nearest integer. That is correct! 14 dog heights$$ 14 dog heights - correct Answer Explanation Correct answers: 10 847 st u d e n t s - incorrect Answer Explanation 11 Correct answers: • 846 st u d e n t s $846\ students$ • Given the information in the question, EBP=0.04since 4%=0.04and zα2=z0.01=2.326 because the confidence level is 98%. The values of p′a n d q′a r e unknown, but using a value of 0.5for p′w i l l 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 846s t u d e n t s . • • • • Question 6 · 1/1 points The average score of a random sample of 87s e n i o r 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%c onf ide nc e interval for the mean of the population. Round your answers to two decimal places and use ascending order. That is correct! 12 open paren 509 point 3 0 comma 519 point 1 8 close paren$$ open paren 509 point 3 0 comma 519 point 1 8 close paren - correct 15 That's not right. 2 point 8 4 5$$ 2 point 8 4 5 - incorrect Answer Explanation Correct answers: • 0 point 0 2 6 $0.026$ • Given the population proportion p=8%=0.08and a sample size of n=110, the standard deviation of the sampling distribution of sample proportions is σp̂=p(1−p)n‾‾‾‾‾‾‾‾‾√=0.08(1−0.08)110‾‾‾‾‾‾‾‾‾‾‾‾‾‾√≈0.026 • • • • Question 21 · 1/1 points In order to estimate the average electricity usage per month, a sample of 125re s i de n t i a l 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% 16 confidence interval for the mean electricity usage in kilowatt hours. Round your answers to two decimal places and use ascending order. 17 That is correct! open paren 894 point 4 3 comma 940 point 2 1 close paren$$ open paren 894 point 4 3 comma 940 point 2 1 close paren - correct Answer Explanation Correct answers: • open paren 894 point 4 3 comma 940 point 2 1 close paren $\left(894.43,\ 940.21\right)$ • A 98%confidence interval for μis (x̄−zα/2σn‾√,x̄+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, x̄, using the AVERAGE function. Thus, the sample mean is x̄=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‾√≈2.89. The confidence interval for the population mean has a lower limit of 917.32−22.89=894.43and an upper limit of 917.32+22.89=940.21. Thus, the 98%confidence interval for μis (894.43, 940.21). • • 20 Suppose Hugo types 69w o r d s per minute in a typing test on Wednesday. The z-score when x=69is 0.333. This z-score tells you that x=69is 0.333standard deviations to the right of the mean, 64. 21 Suppose Hugo types 69w o r d s per minute in a typing test on Wednesday. The z-score when x=69is 0.417. This z-score tells you that x=69is 0.417standard deviations to the right of the mean, 64. Answer Explanation Correct answer: Suppose Hugo types 69w o r d s per minute in a typing test on Wednesday. The z-score when x=69is 0.417. This z-score tells you that x=69is 0.417standard deviations to the right of the mean, 64. The z-score can be found using the formula z=x−μσ=69−6412=512≈0.417 A positive value of zmeans that that the value is above (or to the right of) the mean, which was given in the problem as μ=64words per minute in a typing test. The z-score tells you how many standard deviations the value xi s above (to the right of) or below (to the left of) the mean, μ. So, typing 69wor d s per minute is 0.417standard deviations away from the mean. Your answer: Suppose Hugo types 69w o r d s per minute in a typing test on Wednesday. The z-score when x=69is 0.333. This z-score tells you that x=69is 0.333standard deviations to the right of the mean, 64. • • • • 22 Question 23 25 Suppose Hugo types 65w o r d s per minute in a typing test on Wednesday. The z-score when x=65is 2.471. This z-score tells you that x=65is 2.471standard deviations to the right of the mean, 71. Suppose Hugo types 65w o r d s per minute in a typing test on Wednesday. The z-score when x=65is −1.091. This z-score tells you that x=65is 1.091standard deviations to the left of the mean, 71. Answer Explanation Correct answer: Suppose Hugo types 65w o r d s per minute in a typing test on Wednesday. The z-score when x=65is −1.091. This z-score tells you that x=65is 1.091standard deviations to the left of the mean, 71. The z-score can be found using the formula z=x−μσ=65−715.5=−65.5≈−1.091 A negative value of zme ans that that the value is below (or to the left of) the mean, which was given in the problem as μ=71words per minute in a typing test. The z-score tells you how many standard deviations the value xi s above (to the right of) or below (to the left of) the mean, μ. So, typing 65wor d s per minute is 1.091standard deviations away from the mean. Your answer: Suppose Hugo types 65 words per minute in a typing test on Wednesday. The z-score when x = 65 is \mathbf{-2.471}. This z-score tells you that x = 65 is \mathbf{2.471} standard deviations to the left of the mean, \mathbf{71}. • 26 • • • 27 Question 24 · 1/1 points 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 $_136$_ pages if the mean is $_194$_ pages and the standard deviation is $_29$_ pages? Use the empirical rule.Enter your answer as a percent rounded to two decimal places if necessary. That is correct! 2 point 5 percent$$ 2 point 5 percent - correct Answer Explanation Correct answers: • 2 point 5 percent $2.5\%$ • Notice that $_136$_ 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. • • • •