MATH 225N Week 6 quiz (2020) – Chamberlain College of Nursing | MATH225N Week 6 quiz (2020) – Graded A

MATH 225N Week 6 quiz (2020) – Chamberlain College of Nursing 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 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! $$(67, 87) Answer Explanation Correct answers: • $left(67, 87right)$(67, 87) A confidence interval is an interval of values, centered on a point estimate, of the form (pointestimate−marginof error,pointestimate+marginof error) Using the given point estimate for the mean, x¯=77 and margin of error 10, the confidence interval is: (77−10,77+10)(67,87) QUESTION 2 1/1 POINTS A random sample of adults were asked whether they prefer reading an e-book over a printed book. The survey resulted in a sample proportion of p′=0.14, with a sampling standard deviation of σp′=0.02, who preferred reading an e-book.Use the empirical rule to construct a 95% confidence interval for the true proportion of adults who prefer e-books. That is correct! $$(0.10, 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′=0.14 and 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) 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% 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. QUESTION 4 1/1 POINTS 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! $$14 dog heights Answer Explanation Correct answers: • $14text{ dog heights}$14 dog heights The formula for sample size is n=z2σ2EBM2. In this formula, z=zα2=z0.025=1.96, because the confidence level is 95%. From the problem, we know that σ=3.7 and EBM=2. Therefore, n=z2σ2EBM2=(1.96)2(3.7)222≈13.15. Use n=14 to ensure that the sample size is large enough. Also, the sample size formula shown above is sometimes written using an alternate format of n=(zσE)2. In this formula, E is used to denote margin of error and the entire parentheses is raised to the exponent 2.Therefore, the margin of error for the mean can be denoted by "EBM" or by "E". Either formula for the sample size can be used and these formulas are considered as equivalent. QUESTION 5 1/1 POINTS Clarence wants to estimate the percentage of students who live more than three miles from the school. He wants to create a 98% confidence interval which has an error bound of at most 4%. How many students should be polled to create the confidence interval? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 Use the table of values above. That is correct! $$846 students Answer Explanation- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Answer Explanation Correct answer: Suppose Hugo types 76 words per minute in a typing test on Wednesday. The z-score when x=76 is 2.667. This z-score tells you that x=76 is 2.667 standard deviations to the right of the mean, 60. The z-score can be found using the formula z=x−μσ=76−606=166≈2.667 A positive value of z means that that the value is above (or to the right of) the mean, which was given in the problem as μ=60 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 76 words per minute is 2.667 standard deviations away from the mean. QUESTION 23 1/1 POINTS Hugo averages 51 words per minute on a typing test with a standard deviation of 14 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(51,14). Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is ________. This z-score tells you that x=66 is ________ standard deviations to the ________ (right/left) of the mean, ________. Correctly fill in the blanks in the statement above.That is correct! Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is −1.071. This z-score tells you that x=66 is 1.071 standard deviations to the left of the mean, 51. Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is 1.071. This z-score tells you that x=66 is 1.071 standard deviations to the right of the mean, 51. Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is −0.882. This z-score tells you that x=66 is 0.882 standard deviations to the left of the mean, 51. Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is 0.882. This z-score tells you that x=66 is 0.882 standard deviations to the right of the mean, 51. Answer Explanation Correct answer: Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is 1.071. This z-score tells you that x=66 is 1.071 standard deviations to the right of the mean, 51. The z-score can be found using the formula z=x−μσ=66−5114=1514≈1.071 A positive value of z means that that the value is above (or to the right of) the mean, which was given in the problem as μ=51 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 66 words per minute is 1.071 standard deviations away from the mean. 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 fewerthan 133 pages if the mean is 189 pages and the standard deviation is 28 pages? Use the empirical rule.Enter your answer as a percent rounded to two decimal places if necessary. That is correct! $$2.28% Answer Explanation Correct answers: • $2.5%$2.5% Notice that 133 pages is two standard deviations less than the mean. Based on the empirical rule, 95% of the number of pages in the books are within two standard deviations of the mean. Since the normal distribution is symmetric, this implies that 2.5% of the number of pages in the books are less than two standard deviations less than the mean. QUESTION 25 1/1 POINTS Lisa has collected data to find that the number of pages per book on a book shelf has a normal distribution. What is the probability that a randomly selected book has fewer than 142 pages if the mean is 190 pages and the standard deviation is 24 pages? Use the empirical rule.Enter your answer as a percent rounded to two decimal places if necessary. That is correct! $$2.28% 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 standarddeviations 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.