72) Express the confidence interval (0.432, 0.52) in the form of p± E. A) 0.432 ± 0.088 B) 0.432 + 0.044 C) 0.476 ± 0.088 D) 0.476 ± 0.044
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- Let X1,..., X100 ~ Bernoulli(p) be iid, where 0Express the confidence interval 23.5±101.123.5±101.1 in the form of a trilinear inequality. <μ<Give a 98% confidence interval, for μ1−μ2 given the following information. n1=45 xbar1=2.62, s1=0.8n2=60n xbar = 3.02, s2=0.67 ____ ±± ____ Use Technology Rounded to 2 decimal places.Which of the following defines the 90% confidence interval when σ is known and n = 25? a) m±1.711(σm) b) m±2.797(σm) c) m±2.33(σm) d) m±1.645(σm) e) m±2.58(σm) The correct answer is d: m±1.645(σm) Please provide a solution to the question and explain.Give a 90% confidence interval, for μ1−μ2μ1-μ2 given the following information. n1=20, x¯1=2.95, s1=0.7n2=30, ¯x2=2.76 , s2=0.75 __±___ Use Technology Rounded to 2 decimal places.A 95% confidence interval for the mean μ of a population is computed from a random sample and found to be 10±4. We may conclude? 1) that there is a 95% probability that the true mean is 10 and a 95% chance the true margin of error is 4. 2) that there is a 95% probability that μ is between 6 and 14. 3) 95% of the population is between 6 and 14. 4) that if we took many, many additional samples and from each computed a 95% confidence interval for μ, approximately 95% of these intervals would contain μ. 5) All of the above.Construct a 95% confidence interval to estimate the population mean when x = 54 and s = 10.1 for the sample sizes below. a) n=22 b) n= 43 c) n= 60 ... a) The 95% confidence interval for the population mean when n= 22 is from a lower limit of i to an upper limit of (Round to two decimal places as needed.)Construct the confidence interval for the population mean μ c = 0.98, x̅ = 8.6, σ = 0.3, n = 59 A 98% cofidence interval for μ is ([ ], [ ]) (round to two decimal places as needed.)The life time (age) in hours of a random sample of one of the batteries produced in Jordan gave the following summary: of Sample size Sample Average Sample Standard deviation n = 9 X = 95 S = 3 A 98% confidence interval for the population standard deviation O is: a) (1.09 , 17.86) b) (1.09 , 6.61) c) (1.89 , 6.61) d) (3.58 , 43.7) e) (1.04, 4.23)For a one-tailed test (lower tail) with n = 10 at 90% confidence (i.e. α = 0.10), the appropriate tα (critical value) is a. -1.383 b. 2.821 c. 1.383 d. -2.821Express the confidence interval 62.3%<p<78.3%62.3%<p<78.3% in the form of ˆp±Ep^±E.% ±± %Express the confidence interval (106.2,250)(106.2,250) in the form of ¯x±MEx¯±ME.¯x±ME=x¯±ME= ±±SEE MORE QUESTIONS