If all other quantities remain the same, how does the indicated change affect the width of a confidence interval? (a) Increase in the level of confidence (b) Increase in the sample size (c) Increase in the population standard deviation (a) How does an increase in the level of confidence affect the width of a confidence interval? Choose the correct answer below. An increase in the level of confidence V the confidence interval.
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- A survey of 50 young professionals found that they spent an average of $24.15w hen dining out, with a standard deviation of $12.06. Can you conclude statistically that the population mean is greater than $27? Use a 95% confidence interval is. The 95% confidence interval is _____ , _______As $27 is__________ of the confidence interval, we________ , conclude that the population mean is greater than $27. (Use ascending order. Round to four decimal places as needed.)Answer all the questions asked. Starting from Employed in field of study to prefer to work for a large companyDetermine the confidence level. Determine the confidence level for confidence interval shown in the figure below. 5-10 =.10 98% 95% 90% 80% arch Cc F2 F3 F4 F5 F6 F7 F8 F9 F10 & 4. 5 * 00 %#3
- W: 6.1 & 6.2 16 K Question 14, 6.1.35 Part 1 of 3 HW Score: 47.14%, 16.5 of 35 points O Points: 0 of 1 Save You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 55 home theater systems has a mean price of $119.00. Assume the population standard deviation is $16.30. Construct a 90% confidence interval for the population mean. The 90% confidence interval is (D). (Round to two decimal places as needed.) e solve this View an example Get more help - sty cloudy N 40+ # Q Search % 0+ A 0 6 7 28 & 8 9 0 Clear all Delete Backspace Lock 10.00 PM 4372024 7 Home CO 6 PgUpConstructing a Confidence Interval for a Population Mean A biologist measures the lengths of a random sample 30 mature brown trout in a large lake and finds that the sample a mean weight of 41 pounds. Assume the population standard deviation is 3.7 pounds. Based on this, construct a 90% confidence interval for the mean weight of all mature brown trout in the lake. Round your anwers to two decimal places.Solve it correctly please. I
- INFERENCE FOR SINGLE POPULATION PROPORTION (CONFIDENCE INTERVAL) 11. A random sample of 400 electronic components manufactured by a certain process is tested, and 30 are found to be defective. i. Let p be the proportion of defective components. Find a 95% confidence interval for p. ii. How many components must be sampled so that the 95% confidence interval will specify the proportion of defective components to within ±0.02? Source: Statistics for Engineers and Scientists, 3rd Edition, William, C. Navidi.Statistics helpLibrary A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 90% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi? 0.58 0.74 0.10 0.92 1.32 0.49 0.96 t: 0 What is the confidence interval estimate of the population mean p? ppmUSE 3 DECIMALS A researcher believes that reading habit of newspaper is decreasing every year. In order to see the true population proportion of people who are reading newspaper s/he collect a randomly selected people on the yearly basis. The following table gives the number of people who do not read newspaper with respect to the years. Construct a 90 % confidence interval for the true population proportion difference of people who are reading newspaper in 2005 and 2010. (P2005 - P2015) 2010 2005 In 1200 1000 Number of people who do not read newspaper 300 400 KOncekiAges of students: A simple random sample of 70 U.S. college students had a mean age of 21.53 years. Assume the population standard deviation is o - 3.82 years. Construct an 80% confidence interval for the mean age of U.S. college students. Round the answers to two decimal places. An 80% confidence interval for the mean age of U.S. college students isUSE 3 DECIMAL A researcher believes that reading habit of newspaper is decreasing every year. In order to see the true population proportion of people who are reading newspaper s/he collect a randomly selected people on the yearly basis. The following table gives the number of people who do not read newspaper with respect to the years. Construct a 90 % confidence interval for the true population proportion difference of people who are reading newspaper in 2005 and 2010. 20102005 n 12001000 Number of people who do not read newspaper 300 400SEE MORE QUESTIONS