Given α = 0.01, x²=X0.005 = 49.645, s² = 213.16, and n = 28 compute the lower bound for a 99% confidence interval using the formula s(n-1) LB = (Round the answer to 2 decimal places.)
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- 2.4 Compute the 95% confidence interval for the population (true) value of prediction computed in question 2.3, and interpret it in words as well as graphically. The formula for the 95% confidence interval around a predicted Y value is: Y= αzerit(a=.05, 2 tailed)*sy.x Define and give the values for the following the terms in the formula. Î: 4.15504 Zsit: 0.025 Sy-x: 1.593 The numerical values for the confidence interval are:A 100(1 - α)% confidence interval for the mean μ of a normal population when the value of σ is known is given by (x-²a/2" Six+ .za/2 VM). By how much must the sample size n be increased if the width of the CI above is to be halved? Halving the length requires n to be increased by a factor of If the sample size is increased by a factor of 16, what effect will this have on the width of the interval? Increasing the sample size by a factor of 16 will ---Select-- the length by a factor of ---Select-- increase Need Help? Read It decreaseA simple random sample of size n = 40 is drawn from a population. The sample mean is found to be x = 121.2 and the sample standard deviation is found to be s = 12.6. Construct a 99% confidence interval for the population mean. The lower bound is (Round to two decimal places as needed.) The upper bound is . (Round to two decimal places as needed.)
- In random, independent samples of 225 adults and 200 teenagers who watched a certain television show, 112 adults and 138 teens indicated that they liked the show. Let p1 be the proportion of all adults watching the show who liked it, and let p2 be the proportion of all teens watching the show who liked it. Find a 95% confidence interval for −p1p2. Then find the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. a.) The Lower Limit is: b.) The Upper Limit is:If n=16, ¯xx¯(x-bar)=35, and s=17, construct a confidence interval at a 80% confidence level. Assume the data came from a normally distributed population.Give your answers to one decimal place. I used the formula 35-NORM.INV(1-2.0/2,0,1)*((17)/SQRT(16)) for the lower and the same but with a plus sign for the upper, I got it wrong and dont understand why. We have to use Excel so please explain what excel formula to use.A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 108, and the sample standard deviation, s, is found to be 10. (a) Construct a 98% confidence interval about μ if the sample size, n, is 16. (b) Construct a 98% confidence interval about µ if the sample size, n, is 12. (c) Construct a 70% confidence interval about u if the sample size, n, is 16. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Click the icon to view the table of areas under the t-distribution. (a) Construct a 98% confidence interval about μ if the sample size, n, is 16. Lower bound:; Upper bound: (Use ascending order. Round to one decimal place as needed.)
- Solve the given problem using mathematical procedure.(a) A random sample of n = 36 students who took an examination yields the confidence interval (73 , 74), for the population mean u. The standard deviation o = 1.875. Find the confidence coefficient 100(1 – a)% of this interval. (b) Assume o = 2. How many students should be selected in order to be 98% confident that the margin of error E = +0.6 ?- Construct the 99% confidence interval for the difference μ₁ −μ₂ when x₁=476.88, x₂=318.32, s₁=42.11, s₂=26.80, n₁=19, and n₂=22. Use tables to find the critical value and round the answers to at least two decimal places. A 99% confidence interval for the difference in the population means isA simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 109, and the sample standard deviation, s, is found to be 10. (a) Construct a 95% confidence interval about if the sample size, n, is 14. (b) Construct a 95% confidence interval about if the sample size, n, is 18. H (c) Construct a 99% confidence interval about u if the sample size, n, is 14. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Lower bound: Upper bound: (Use ascending order. Round to one decimal place as needed.) COLL SaveIF n < 30 AND T = Ý−µ -H, √n SHOW HOW THE 100(1-2x)% CONFIDENCE INTERVAL FOR U IS DERIVEDA sample of n = 7 scores is selected from a population with an unknown mean ( µ). The sample has a mean of M = 40 and a variance of s 2 = 63. Which of the following is the correct 95% confidence interval for µ?SEE MORE QUESTIONS