Bees. Solve the problem. Essay Example for Free

Bees. Solve the problem. EssaySolve the problem.1) Find the vital value that corresponds to a degree of corporate trust of 91%. A) 1.70B) 1.34 C) 1.645 D) 1.752) The following self-assertion interval is obtained for a tribe proportion, p0.817 p 0.855 Use these confidence interval limits to find the point estimate, A) 0.839 B) 0.836 C) 0.817 D) 0.833Find the permissiveness of break 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 minimal test coat you should use to assure that your estimate of will be within the required borderline of defect around the population p. 4) circumference of error 0.002 confidence take aim 93% and unkn birth A) 204,757 B) 410 C) 204,750 D) 4055) bank 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 data to arrive at a confid ence interval for the population proportion p.6) When 343 college students are randomly selected and surveyed, it is found that 110 hold 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 humble . 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 incriminate score of 20.2 with a standard exit of 4.6 on a placement 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 examination ExamMean Standard deviation Variance =Mean from a frequency distribution unravel rule of thumbEmpirical 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 distribution binomial 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 val ueMargin of error pSample size p orMargin of error meanSample size meanMargin of error mean