Billy wants to estimate the percentage of students who live more than three miles from the school. He wants to create a95% confidence interval which has a margin of error of at most 3%. How many students should be polled to create theconfidence interval?Z0.10Z0.05 Z0.025 Z0.01 Zo.0051.282 1.645 960 2.326 2.576Use the table of values above.Select the correct answer below:0 203O 267O 7521,068O 2,135

The 95% confidence interval:The 95% confidence interval for the population proportion, π is such…

Answered: Billy wants to estimate the percentage…

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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Use the confidence interval to find the estimated margin of error. Then find the sample mean. A biologist reports a confidence interval of (1.9.3.1) when estimating the mean height (in centimeters) of a sample of seedlings. The estimated margin of error is The sample mean is
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.
If [0.64, 0.82] is a confidence interval for the population proportion. Find the value of the point estimator of the population proportion. Select one: O a. 0.73 O b. 0.70 O c. 0.74 O d. 0.72
A poll of 2338 U.S. adults found that 71% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to %
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Use the confidence interval to find the estimated margin of error. Then find the sample mean. A biologist reports a confidence interval of (2.8,4.6) when estimating the mean height (in centimeters) of a sample of seedlings. The estimated margin of error is The sample mean is
Which of the following would result in the widest confidence interval? A sample size of 30 with 95% confidence. A sample size of 100 with 95% confidence A sample size of 100 with 99% confidence. A sample size of 30 with 99% confidence.
Use the histogram below to build a 95% confidence interval of the average heart rate of a person. Put your two numbers below in the spaces provided. Round to one decimal: ex: 98.2 The histogram visualizes the results of doing one thousand studies just like above. We are 95% confident that the average heart rate of a person is between ____and_____ All information needed to answer this problem is present. There is no missing information to this question.
Use the confidence interval to find the margin of error and the sample mean. (1.57,1.95) The margin of error is enter your response here.
A statistics practitioner took a random sample of 48 observations from a population whose standard deviation is 32 and computed the sample mean to be 107. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 90% confidence. Confidence Interval = C. Estimate the population mean with 99% confidence. Confidence lInterval =
Using the data attached, calculate the confidence interval, and the error bound: 1. How much area is in both tails(combined)? a= 2.How much area is in each tail? a/2 3. Some students think that a 90% confidence interval contains 90% of the data. Use the list of data attached and count how many data values lie within the confidence interval. What percent is this? Is this percent close to 90%? Explain why this percent should or should not be close to 90%?

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