A random sample of 11 health maintenance organizations (HMOS) was selected. For each HMO, the co-payment (in dollars) for a doctor's office visit wasrecorded. The results are as follows.41, 49, 54, 29, 45, 35, 59, 44, 43, 32, 27Send data to calculatorUnder the assumption that co-payment amounts are normally distributed, find a 90% confidence interval for the mean co-payment amount in dollars. Givethe lower limit and upper limit of the 90% confidence interval.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)Lower limit: ||Upper limit:|

The provided information is x x2 41 1681 49 2401 54 2916 29 841 45 2025 35 1225 59…

Answered: A random sample of 11 health…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A recent survey showed that 42% of US employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 500 companies. Find the 99% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage. (Round to four decimal places)
Out of 400 people sampled, 276 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.
The Management Department recently purchased a smallcopier for faculty use. Although the workload of the office staff has improved somewhat, the secretaries are still mak-ing too many trips to the Dean’s office when the depart-mental copier is out of service. Sylvia, the departmental secretary, has been keeping track of failure rates and ser-vice times. With a mean time between failures of 100 hours and a mean time to repair of 24 hours, how much of thetime is the new copier available for faculty use?
During the nutrition research, the amount of consumed kilocalories per day was measured for 18 people - 10 women and 8 men. Results are as follows: Women: 2026, 1600, 1700, 1950, 1922, 1786, 1712, 1777, 2006, 1832; Men: 2170, 2557, 2384, 2427, 2334, 2228, 2161, 2459. Calculate a 90% confidence interval on the mean for women and men separately. Assume distribution to be normal. Round your answers to the nearest integer (e.g. 9876). Women: i Men: i
In a survey of 2567 adults, 1431 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
In a survey of 2024 adults in a recent year, 709 made a New Year's resolution to eat healthier. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.
During the nutrition research, the amount of consumed kilocalories per day was measured for 18 people - 10 women and 8 men. Results are as follows: Women: 1624, 1608, 2003, 1654, 1907, 1636, 1924, 1868, 1578, 2010; Men: 2360, 2547, 2524, 2056, 2172, 2464, 2253, 1954. Calculate a 95% confidence interval on the mean for women and men separately. Assume distribution to be normal. Round your answers to the nearest integer (e.g. 9876). Women: i i Men: i
Please show your work, -)
In a survey of 3500 adults, 1443 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
The number of unhealthy days based on the AQI (air quality index) for a random sample of metropolitan areas is shown. Construct an 80% confidence interval based on the data. Assume the variable is normally distributed. 23, 21, 14, 5, 40, 61, 39, 1, 0, 16
I chose Option 1 for this week’s discussion post. I want to estimate the population mean number of times adults between the ages of 30 and 45 went to the grocery store in the last week. I spoke with 12 adults and asked them how many times they went to the grocery store in the last week. The results were 0, 2, 2, 7, 5, 1, 0, 3, 1, 1, 4, and 3. Now calculate a 95% confidence interval for the population mean. (Round to three decimal places)

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