A researcher plans to use a random sample of n=500 families to estimate the mean monthly family income for a large population. A 99% confidence interval basedon the sample would bethan a 90% confidence interval.narrower and would involve a larger risk of being incorrectwider and would involve a smaller risk of being incorrectwider and would involve a larger risk of being incorrectwider, but it cannot be determined whether the risk of being incorrect would be larger or smallernarrower and would involve a smaller risk of being incorrect

Answered: A researcher plans to use a random…

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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In 2013, the Pew Research Foundation reported that "48% of U.S. adults report that they live with one or more chronic conditions". However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about 1.6%, and a normal model may reasonably be used in this setting. Create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions. Report your answer as a percentage rounded to 3 decimal places. We are 95% confident that the true proportion of U.S. adults who live with one or more chronic conditions is between % and
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Sal knows that parties of people on average spend 75 minutes at their restaurant. They wonder if this average is significantly different on the weekends. They found the average time spent by the 30 parties which dined there this weekend was 82 minutes with a SD of 17 minutes. Assume an alpha of .05. The lower end of the 95% confidence interval is and the higher end of the 95% confidence interval is.
One company recently conducted a survey in which 420 women and 420 men were asked "Are you satisfied with the work of the Minister of Health?". Of these, 155 men and 118 women said they were satisfied with the minister's work. Assume that the experimental design was correct. Questions 1 and 2 cover this survey. 1. Find the upper limit of 95% confidence interval for the difference in the proportion of men and women who are happy with the work of the Minister (let men be group 1). 1. 0.03. 2. 0.09. 3. 0.15. 4. It is not possible to apply a normal approach and therefore we can not calculate the confidence interval. 2. What is the rejection area if it is to be examined whether the proportion of men and women who are satisfied with the job of Prime Minister is different (use a = 0.05)? 1. RejectHO efz 1.96. 2. Reject HO if z> 1.64. 3. RejectHO efz> -1.96 or <1.96. 4. Insufficient information provided to answer this.
A researcher wants to estimate the true proportion of Americans who suffer side-effects after taking a particular medication. A 99% confidence interval for the true proportion of Americans who suffer side-effects after taking the particular medication was found to be (0.024, 0.160). If a hypothesis test is conducted with the hypotheses, H,: p = 0.10 vs H: p# 0.10 , what would be the most likely conclusion? Select one: a. There is very strong evidence that p 0.10. Ob. There is little to no evidence that p = 0.10. c. There is little to no evidence that p # 0.10. d. Cannot tell.
I'm having trouble trying to figure out how to work the problem below out. Let’s say that 44 percent of the 1,300 U.S. adults who responded to a poll about marijuana legalization said they support legalization. Let’s also say the margin of error for the results is +/- 3 percentage points. The margin of error tells us there’s a high probability that nationwide support for marijuana legalization falls: The correct interpretation of this confidence interval is:
When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. 3. a. Find a 99% confidence interval estimate of the percentage of yellow peas.
At a charter high school, the average GPA of a high school senior is known to be normally distributed. After a sample of 27 seniors is taken, the average GPA is found to be 2.85 and the sample variance is determined to be 0.16. Find a 95% confidence interval for the population mean GPA. a. 2.5615 to 2.8585 b. 2.6358 to 3.0642 c. 2.6917 to 3.0083 d. 2.5167 to 2.9033
Suppose the goal is to estimate the parameter stated in question 3 using a 95% confidence interval with margin of error no larger than 0.048. What is the minimum number of trick-or-treaters that would need to be selected to allow the calculation of a 95% confidence interval with margin of error no larger than 0.048? It can be assumed for this problem only that the proportion of all people who went trick-or-treating who received fruit in addition to candy is 0.24 and this number can be used for this problem. Which of the following is the correct answer? 204.13 123.14 304.13 123 305 205 304 124 204
Suppose a sample of 9 customers spends an average of $200 with a standard deviation $20 what is a 95% confidence interval for the sample mean. Use the fact that 95% of sample means are within 2.306 standard errors of the sample mean.
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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