Week 6 Math 225N Statistics Quiz_Fall 2020_Already Graded A.

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. ...

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! $$(67, 87) Answer Explanation Correct answers:  $\left(67,\ 87\right)$(67, 87) A confidence interval is an interval of values, centered on a point estimate, of the form (pointestimate−marginof error,pointestimate+marginof error) Using the given point estimate for the mean, x¯=77 and margin of error 10, the confidence interval is: (77−10,77+10)(67,87) QUESTION 2 1/1 POINTS A random sample of adults were asked whether they prefer reading an e-book over a printed book. The survey resulted in a sample proportion of p′=0.14, with a sampling standard deviation of σp′=0.02, who preferred reading an e-book. Use the empirical rule to construct a 95% confidence interval for the true proportion of adults who prefer e-books.That is correct! $$(0.10, 0.18) Answer Explanation Correct answers:  $\left(0.10,\ 0.18\right)$(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! $$14 dog heights Answer Explanation Correct answers:  $14\text{ dog heights}$14 dog heights The formula for sample size is n=z2σ2EBM2. In this formula, z=zα2=z0.025=1.96, because the confidence level is 95%. From the problem, we know that σ=3.7 and EBM=2. Therefore, n=z2σ2EBM2=(1.96)2(3.7)222≈13.15. Use n=14 to ensure that the sample size is large enough. Also, the sample size formula shown above is sometimes written using an alternate format of n=(zσE)2. In this formula, E is used to denote margin of error and the entire parentheses is raised to the exponent 2.Therefore, the margin of error for the mean can be denoted by "EBM" or by "E". Either formula for the sample size can be used and these formulas are considered as equivalent. QUESTION 5 1/1 POINTS Clarence wants to estimate the percentage of students who live more than three miles from the school. He wants to create a 98% confidence interval which has an error bound of at most 4%. How many students should be polled to create the confidence interval? z0.10 z0.05 z0.02 5 z0.01 z0.005 1.28 2 1.64 5 1.96 0 2.32 6 2.57 6 Use the table of values above. That is correct! $$846 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 846 students. QUESTION 6.