Bees. Solve the problem. Essay Example for Free

Bees. Solve the problem. EssaySolve the problem.1) Find the critical value that corresponds to a degree of confidence of 91%. A) 1.70B) 1.34 C) 1.645 D) 1.752) The following confidence breakup is obtained for a population proportion, p0.817 p 0.855 Use these confidence interval limits to find the take aim adjudicate, A) 0.839 B) 0.836 C) 0.817 D) 0.833Find the margin of error for the 95% confidence interval used to estimate the population proportion. 3) n = 186, x = 103A) 0.0643 B) 0.125 C) 0.00260 D) 0.0714Find the nominal sample size of it you should use to assure that your estimate of will be within the required margin of error around the population p. 4) Margin of error 0.002 confidence train 93% and unknown A) 204,757 B) 410 C) 204,750 D) 4055) Margin of error 0.07 confidence level 95% from a prior study, is estimated by the decimal equivalent of 92%.A) 58 B) 174 C) 51 D) 4Use the given degree of confidence and sample selective information to construct a confidence interval for the population proportion p.6) When 343 college students are indiscriminately selected and surveyed, it is found that 110 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car. A) 0.256 p 0.386 B) 0.279 p 0.362C) 0.271 p 0.370 D) 0.262 p 0.379Determine whether the given conditions justify using the margin of error E = when finding a confidence interval estimate of the population mean . 7) The sample size is n = 9, is not known, and the original population is normally distributed. A) Yes B) NoUse the confidence level and sample data to find the margin of error E. 8) Systolic blood pressures for women aged 18-24 94% confidence n = 92, x = 114.9 mm Hg, = 13.2 mm HgA) 47.6 mm Hg B) 2.3 mm Hg C) 2.6 mm Hg D) 9.6 mm HgUse the confidence level and sample data to find a confidence interval for estimating the population .9) A group of 52 randomly selected students have a mean score of 20.2 with a standard deviation of 4. 6 on a stead test. What is the 90 percent confidence interval for the mean score, , of all students taking the test?A) 19.1 21.3 B) 18.7 21.7C) 19.0 21.5 D) 18.6 21.8Use the margin of error, confidence level, and standard deviation to find the minimum sample size required to estimate an unknown population mean . 10) Margin of error $100, confidence level 95%, = $403A) 91 B) 63 C) 108 D) 44Formula sheet for Final ExamMean Standard deviation Variance =Mean from a frequency dispersion Range rule of alternateEmpirical Rule 68-95-99.7 z score weighted meanOutliersif A and B are mutually exclusiveif A and B are not mutually exclusiveif A and B are independentif A and B are dependentComplementary eventsmean of a probability distributionstandard deviation of aprobability distributionBinomial probabilityBinomial probability calculatorExactly binompdf(n,p,x)At least 1 binomcdf(n,p,x 1)At most binomcdf(n,p,x)Binomial meanBinomial standard deviationExpected valueMargin of error pSa mple size p orMargin of error meanSample size meanMargin of error mean