A survey of 1000 air travelers¹ found that 60 % prefer a window seat. The sample size is large enough to use the normal distribution,and a bootstrap distribution shows that the standard error is SE = 0.015.Use a normal distribution to find a 99% confidence interval for the proportion of air travelers who prefer a window seat.Round your answers to three decimal places.The 99% confidence interval is i 0.621to0.698

The sample size is 1000, the sample proportion is 0.60 and standard error is 0.015.

Answered: A survey of 1000 air travelers¹ found…

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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IMR in Singapore. The infant mortality rate (IMR) is the number of infant deaths per 1000 live births. Suppose that you have been commissioned to estimate the IMR in Singapore. From a random sample of 1109 live births in Singapore, you find that 0.361% of them resulted in infant deaths. You next find a 90% confidence interval: or 0.000647 to 0.00657. You then conclude, “I can be 90% confident that the IMR in Singapore is somewhere between 0.647 and 6.57.” How did you do?
Tina catches a 14-pound bass. She does not know the population mean or standard deviation. So she takes a sample of five friends and they say the last bass they caught was 9, 12, 13, 10, and 10 pounds. Find the t and calculate a 95% (α = .05) confidence interval.
Construct the 99.5% confidence interval for the difference p, - p, when x, = 50, n, = 75, x, = 65, and n, = 151. Round the answers to three decimal places. A 99,5% confidence interval for the difference between the two proportions is
A statistics class is estimating the mean height of all female students at their college. They collect a random sample of 42 female students and measure their heights. The mean of the sample is 65.3 inches and the sample standard deviation is 5.4 inches. Find the 90% confidence interval for the mean height of all female students in their school? Assume that the distribution of individual female heights at this school is approximately normal. 1. The t value is 1.3025 2. The standard error has a value of -- 3. The margin of error has a value of 4. Find the 90% confidence interval for the mean height of all female students in their school
Echinacea is widely used as an herbal remedy for the common cold, but does it work? In a double-blind experiment, healthy volunteers agreed to be exposed to common-cold-causing rhinovirus type 39 and have their symptoms monitored. The volunteers were randomly assigned to take either a placebo or an echinacea supplement daily for 5 days following viral exposure. Among the 103 subjects taking a placebo, 88 developed a cold, whereas 44 of the 48 subjects taking echinacea 1 developed a cold.
50 of the 600 medical students who were chosen at random stated they intended to work in a rural area. Calculate the 90% confidence interval for the actual percentage of medical students who intend to practice in rural areas. The responses shown here are estimates.
A survey of 32,782 adults in 2004 indicated that 24.6% of adults were current smokers. A similar study conducted in 1995 of 40,000 adults indicated that 29.7% were current smokers. a. Find and interpret a point estimate of the difference between the proportion of current smokers in 1995, and the proportion of current smokers in 2004. b. A 95% confidence interval for the true difference is (-0.058,-0.045). Interpret. c. What assumptions must you make for the confidence interval i part b to be valid? (…) a. Find a point estimate of the difference between the proportion of current smokers in 1995 and the proportion of current smokers in 2004. (Round to three decimal places as needed.)
Use the Student's t distribution to find tc for a 0.95 confidence level when the sample is 29. (Round your answer to three decimal places.)
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What if we're able to use a normal model. To construct a 95% Confidence Interval, we use the following formula: p hat ± ME for 95% CI If p hat = 0.38 and ME for 95% CI = 0.096, what is CI? Keep the numbers in 3 decimal places. Put the two numbers in ( ) and use a comma, then a space between the two numbers.

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