A confidence interval for a population mean has length 20. a) Determine the margin of error. b) If the sample mean is 58.1, obtain the confidence interval. Confidence interval: ( ).
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A: Given confidence interval is (4.6 , 5.2)
Q: Use the confidence interval to find the estimated margin of error. Then find the sample mean. A…
A:
Q: A confidence interval for a population mean has length 26. a) Determine the margin of error. b) If…
A: a) Given Length (l)= 26 Margin of error (E)= length/2 = 26/2= 13 E=13
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A: It is given that, the confidence interval is (4.4, 5.4).
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A: Lower limit = 1.56 Upper limit = 1.96
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A: Given Data Lower limit = 13.7 upper limit = 20.7
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A: Given, x¯ = 5.7σ = 0.3n = 43confidence level α = 0.05 Critical value Zα2 = 1.96
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A: The confidence interval is CI =(LB,UB) CI = (1.53,2.03)
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A: Given95% confidence interval N=361 x = 59.1 seconds. z= 4.9 SecondsFind the margin of error?
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A: Given: confidence interval ( 45.3 , 82.3 ) Need to find sample mean(x̄) and margin of error(ME) .
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A: From the provided information, Confidence interval (1.70, 2.04)
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Q: Use the given confidence interval to find the margin of error and the sample mean. (6.23,10.55) The…
A: Solution: From the given information, the confidence interval is (6.23, 10.55).
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A: It is given that, the confidence interval is (42.6, 78.2).
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- Confidence interval of the population mean is given by: X+2 x where X is the mean, s is the standard deviation and n is the sample size. Create a function called "summary" that lists mean, %15 trimmed mean and the 95% bounds of confidence interval for a random sample. Use the function to calculate the statistics for the given sample where x= (1200, NA, 1205, 1203, 1296,1304) 1. 2.Assume that population mean is to be estimated from the sample described. Use the sample results to approximate the margin of error and 95% confidence interval. n = 36, x = 64.4 seconds, s = 6.5 seconds. The margin of error is seconds. (Round to one decimal place as needed.) an example 4Use technology to construct the confidence intervals for the population variance of and the population standard deviation o. Assume the sample is taken from a normally distributed population. c=0.95, s=37, n= 18 The confidence interval for the population variance is (770 86, 3076.74). (Round to two decimal places as needed.) .... The confidence interval for the population standard deviation is (Round to two decimal places as needed.) More Help me solve this View an example Get more help- Clear all 55°F AO Type here to search 99+ Lenovo AL BTY 10/05/17 Insert Delete PrtSc F11 F12 F9 F10 C F5 F7 F8 F6 F4 F3 Esc F1 F2 Bac @ 23 7 6- 3 4 5 团 %24
- Use the given confidence interval to find the margin of error and the sample mean. (15.3,23.7) The sample mean is . (Type an integer or a decimal.) The margin of error is. (Type an integer or a decimal.)Solve the last partUse the given confidence interval to find the margin of error and the sample mean. (11.3,21.1) The sample mean is The margin of error is
- I collect a random sample of size n from a population and from the data collected compute a 90% confidence interval for the mean of the population. Which of the following would produce a wider confidence interval, based on these same data? Use a larger confidence level. Use a smaller confidence level. a b.7Use the confidence interval to find the estimated margin of error. Then find the sample mean. A store manager reports a confidence interval of (43.1,81.9) when estimating the mean price (in dollars) for the population of textbooks. The estimated margin of error is (Type an integer or a decimal.) The sample mean is (Type an integer or a decimal.)