Use the confidence interval to find the margin of error and the sample mean. (0.196,0.300) The margin of error is
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Q: Use the confidence interval to find the margin of error and the sample mean. (1.56,1.96) The margin…
A: Lower limit = 1.56 Upper limit = 1.96
Q: Use the given confidence interval to find the margin of error and the sample mean. (13.3,22.9)
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A: Givens=4n=26c=0.90
Q: he given confidence interval to find the margin of error and the sample mean. (66.9 ,81.9 )
A: The sample mean can be calculated as: =LowerLimit+UpperLimit2=66.9+81.92=74.4 The sample mean is…
Q: Use the given confidence interval to find the margin of error and the sample mean. (70.7,85.7)…
A: Use the given confidence interval to find the margin of error and the sample mean. (70.7,85.7)…
Q: Use the confidence interval to find the estimated margin of error. Then find the sample mean. A…
A: Confidence interval is (2.5 ,3.1) Lower confidence Limit(LCL)=2.5 Upper Confidence Limit(UCL)=3.1
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Q: Use the confidence interval to find the margin of error and the sample mean. (1.53,2.03) The margin…
A: The confidence interval is CI =(LB,UB) CI = (1.53,2.03)
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A: Solution: 23. From the given information, n=6 and the confidence level is 0.95.
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Q: Use the given confidence interval to find the margin of error and the sample mean. (15.3,22.3)
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A: Given Confidence interval is LL=1.6 UL=2.8 ME=?
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A: From the provided information, Confidence interval (1.70, 2.04)
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A: Confidence interval is (4.63,8.15)
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Q: Use the confidence interval to find the estimated margin of error. Then find the sample mean.
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Q: Use the confidence interval to find the margin of error and the sample mean. (1.48,2.08) The margin…
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- Rosetta wants to estimate the percentage of people who rent their home. She surveys 250 individuals and finds that 48 rent their home. Find the confidence interval for the population proportion with a 90% confidence level. Round the final answer to three decimal places. C.I. = ( , )Use the given confidence interval to find the margin of error and the sample mean. (70.5,84.7)Use the confidence interval to find the margin of error and the sample mean. (1.59,1.95) The margin of error is (Round to two decimal places as needed.)
- Use the confidence interval to find the margin of error and the sample mean. (1.54,1.88) The margin of error is 0.17. (Round to two decimal places as needed.) The sample mean is. (Type an integer or a decimal.)n= 1655 (nubers of tries), x=176 (number of success tries), 98% confidence. Find the confidence interval estimate of the population proportion p.Construct a confidence interval for the population mean. Assume that the population has a normal distribution. n= 30, x bar = 84.6, s=10.5, 90% confidence.
- Use the confidence interval to find the margin of error and the sample mean. (0.346,0.450) The margin of error is The sample mean is 9 五 0 SF10Bert computes a 95% confidence interval for p and obtain the interval [0.600,0.700]. Bert boss says, Give me a 95% confidence interval for p. Calculate the answer for Bert.A poll of 2612 U.S. adults found that 15% 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 ________% 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 99% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: ________% to ______% The more error we allow, the less precise our estimate. Therefore, as the confidence…
- Use the confidence interval to find the estimated margin of error. Then find the sample mean. A biologist reports a confidence interval of (4.2,5.4) when estimating the mean height (in centimeters) of a sample of seedlings. The estimated margin of error is The sample mean is Search 6 17 + & 87 lyn 8 9 DYWAN DDI 8 0 0 0statistics practitioner took a random sample of 48 observations from a population whose standard deviation is 23 and computed the sample mean to be 100. 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 95% confidence, changing the population standard deviation to 47; Confidence Interval = C. Estimate the population mean with 95% confidence, changing the population standard deviation to 10; Confidence Interval =What is the P(t > -1.8) with a sample size of 10.
