Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

The concepts of confidence intervals and hypothesis testing, which are fundamental statistical techniques used to make inferences about population parameters based on sample data. Explanations and examples related to constructing confidence intervals for population means and proportions, as well as conducting hypothesis tests to determine if there is sufficient evidence to support a claim about a population parameter. The content is likely suitable for students in statistics, data analysis, or quantitative methods courses at the university level. The document could be useful as study notes, lecture notes, or for preparing assignments, exams, or other academic exercises related to statistical inference.

Typology: Exams

2023/2024

1 / 28

Download Confidence Intervals and Hypothesis Testing 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 100points was a x̄=77, with a margin of error of 10. Construct a confidence interval for the mean score (out of 100points) on the final exam. That is correct! open paren 67 comma 87 close paren$$ open paren 67 comma 87 close paren - correct Answer Explanation Correct answers: • open paren 67 comma 87 close paren $\left(67,\ 87\right)$ • A confidence interval is an interval of values, centered on a point estimate, of the form (pointestimate−marginoferor,pointestimate+marginoferor) Using the given point estimate for the mean, x̄=77and margin of error 10, the confidence interval is: 2 (77−10,77+10)(67,87) • • • • 5 • Question 3 · 1/1 points The pages per book in a library are normally distributed with an unknown population mean. A random sample of books is taken and results in a 95%confidence interval of (237,293)pages. What is the correct interpretation of the 95%confidence interval? That is correct! We estimate with 95%confidence that the sample mean is between 237and 293pages. We estimate that 95%of the time a book is selected, there will be between 237and 293pages. We estimate with 95%confidence that the true population mean is between 237and 293 pages. 6 Answer Explanation Correct answer: We estimate with 95%confidence that the true population mean is between 237and 293 pages. 7 Once a confidence interval is calculated, the interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated, and state the confidence interval. We estimate with 95%confidence that the true population mean is between 237and 293 pages. • • • • Question 4 · 1/1 points The population standard deviation for the heights of dogs, in inches, in a city is 3.7inches. If we want to be 95%confident that the sample mean is within 2inches of the true population mean, what is the minimum sample size that can be taken? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table above for the z-score, and be sure to round up to the nearest integer. That is correct! 14 dog heights$$ 14 dog heights - correct 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′and q′are unknown, but using a value of 0.5for 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. • • • • Question 6 · 1/1 points The average score of a random sample of 87senior 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. 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 · 16 1/1 points In order to estimate the average electricity usage per month, a sample of 125residential 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. 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‾√≈22.89. 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 69words 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 69words 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 69words 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 xis above (to the right of) or below (to the left of) the mean, μ. So, typing 69words per minute is 0.417standard deviations away from the mean. Your answer: Suppose Hugo types 69words 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. • • • • Question 23 22 · 0/1 points Hugo averages 71words per minute on a typing test with a standard deviation of 5.5words 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(71,5.5). Suppose Hugo types 65words per minute in a typing test on Wednesday. The z-score when x=65is . This z-score tells you that x=65is 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 65words 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 right of the mean, 71. Suppose Hugo types 65words 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 left of the mean, 71. 25 • • 26 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. • • • 27 • Question 25 ·