A confidence interval for a population mean has a margin of error of 2.2. a. Determine the length of the confidence interval. b. If the sample mean is 52.7, obtain the confidence interval. c. Construct a graph that illustrates your results.
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A: Solution: From the given information, confidence level is 0.95, σ=315 and E=105.
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Q: You are given the sample mean and the population standard deviation. Use this information to…
A: sample size(n)=35Mean()=126.00population standard deviation()=$15.20confidence level=90%
Q: You are given the sample mean and the population standard deviation. Use this information to…
A: Given Information: Sample size (n) = 35 Mean price (x) = $ 133 Standard Deviation (s) = $ 15.20
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A: Given thatMargin of error (E) = 4% = 0.04Significance level (α) = 1 - 0.99 = 0.01
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A: GivenMean(x)=145.00standard deviation(σ)=16.60sample size(n)=50
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A: Given n=45, x̄=117, σ=18.30
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Q: technology to construct the confidence intervals. A random sample of 60 home theater systems has a…
A: n = 60 mean = 133 sigma = 17.9
Q: You are given the sample mean and the population standard deviation. Use this information to…
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Q: convenient, use technology to construct the confidence intervals. A random sample of 50 home theater…
A: For the calculation, Minitab is used, We are given a random sample of size 50 with sample mean 126…
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A: Given,n=130X=69p^=Xnp^=69130=0.5311-p^=1-0.531=0.469α=1-0.99=0.01α2=0.005Z0.005=2.576 (from…
Q: You are given the sample mean and the population standard deviation. Use this Info and 95%…
A: sample size, n = 35 mean, x¯ = 131.00 std. dev., σ = 15.50
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Q: Use the confidence interval to find the margin of error and the sample mean. (0.676,0.800)
A: The margin of error is the most common term which we use in a confidence interval. It is the degree…
Q: Use the confidence interval to find the margin of error and the sample mean. (1.57,1.97)
A: The objective of this question is to find the margin of error and the sample mean using the given…
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A: Given data: Sample mean = $145 Sample size (n) = 40 Population standard deviation = $15.50
Q: nstruct a 90% contidence interval for the population mea e 90% confidence interval is ( 139.97…
A: As the population standard deviation is known, we will use z distribution. The critical value for…
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- Help Me please.You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 40 home theater systems has a mean price of $149.00. Assume the population standard deviation is $19.30. Construct a 90% confidence interval for the population mean. The 90% confidence interval is (Round to two decimal places as needed.) Construct a 95% confidence interval for the population mean. The 95% confidence interval is (,). (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. A. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the…You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 35 home theater systems has a mean price of$134.00. Assume the population standard deviation is $18.70. Construct a 90% confidence interval for the population mean. The 90% confidence interval is _____ (Round to two decimal places as needed) Construct a 95% confidence interval for the population mean. The 95% confidence interval is _____ (Round to two decimal places as needed)
- You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 60 home theater systems has a mean price of $125.00. Assume the population standard deviation is $19.30 Construct a 90% confidence interval for the population mean. The 90% confidence interval is___________________________. (Round to two decimal places as needed.)You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 55 home theater systems has a mean price of $117.00.Assume the population standard deviation is $17.40. Construct a 90% confidence interval for the population mean. The 90% confidence interval is (enter your response here,enter your response here). (Round to two decimal places as needed.) Construct a 95% confidence interval for the population mean. The 95% confidence interval is (enter your response here,enter your response here). (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. A. With 90% confidence, it can be said that the population mean price lies in the first…You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals A random sample of 50 home theater systems has a mean price of $125.00. Assume the population standard deviation is $16.40. Construct a 90% confidence interval for the population mean. The 90% confidence interval is ( | ). (Round to two decimal places as needed.) Construct a 95% confidence interval for the population mean. The 95% confidence interval is ( ) (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. O A. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider…
- You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 35 home theater systems has a mean price of $145.00. Assume the population standard deviation is $18.10. Construct a 90% confidence interval for the population mean. The 90% confidence interval is ( D. (Round to two decimal places as needed.) S... el R... 1o-Insta... PDF DRA 2021 Deputy Incidents.pdf Enter your answer in the edit fields and then click Check Answer. 2 parts remaining Clear All Check Answer Riley LAMAYou are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 40 home theater systems has a mean price of $124.00. Assume the population standard deviation is $19.60. Construct a 90% confidence interval for the population mean. The 90% confidence interval is (Round to two decimal places as needed.)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 4
- You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of $142.00. Assume the population standard deviation is $18.80. Construct a 90% confidence interval for the population mean. The 90% confidence interval is (____, ____) (Round to two decimal places as needed.) Construct a 95% confidence interval for the population mean. The 95% confidence interval is (____, ____) (Round to two decimal places as needed.)You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 50 home theater systems has a mean price of $118.00. Assume the population standard deviation is $19.60. Construct a 90% confidence interval for the population mean. The 90% confidence interval is ( ). (Round to two decimal places as needed.) Construct a 95% confidence interval for the population mean. The 95% confidence interval is ( ,). (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. O A. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider…You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 60 home theater systems has a mean price of $110.00. Assume the population standard deviation is $17.20. Construct a 90% confidence interval for the population mean. The 90% confidence interval is ( (Round to two decimal places as needed.)