Essay About Standard Error And Margin Of Error

Investing All Your Money in a Few Randomly Chosen Stocks That Make Up the Index
Ex. 4 a) Investing all your money in a few randomly chosen stocks that make up the index b) Investing all your money in a few randomly chosen stocks that make up the indexEx. 22Assuming proportion as .5 or 50%. P=.5 Standard error = 1.58% Margin of error =2 x Standard error =2*.0158= 3.16%For margin of error as 10% or .1 Standard error =10%/2 =5% .05 =sqrt(.5*.5/n) N=100 The poll sample of 100 is required. For a proportion as 48% or .48 As per my calculations shown below. I would around get 1% or 0.01. This is the Standard Error for a percentage. With a genuine percentage of 48%, The Standard Error for Percentage gives 0.01 or 1%.48% x(100%-48%) = 0.2496.0.2496/2500 =0.00009984SE = SQ(0.00009984) = 0.01ME = 0.01 X 2 = 2%Ex. 28Margin of error =.5% =.005 Standard deviation =2.5% =.025 Assuming the confidence interval as 95% =1.96. (1.96 x .025/ sqrt)/0.05(0.98)2 = 0.9604= 1Therefore, required sample size is 1.Ex. 17Proportion =145/1100 =.1318 Standard error=sqrt(.1318 x (1-.1318)/1100)=.0102 Margin of error =.0102 x 1.96 = .02 Therefore, Lower limit =.1318-.02 =.1118=11.18% Upper limit =.1318+.02 =.1518=15.18% b) 95% sure that between 11.18% to 15.18% workers at a company accessed Internet sites that were clearly not identified with their occupations in the most recent week.c) 10% is excluded in the confidence interval (11.18%-15.18%) consequently the organization statistics is not the same as the national insights.  d) Yes, 95% of individuals at this organization spent in the scope of 4.8 hours give or take 4.2 hours on non-work related destinations on the grounds that the 95% certainty interim is around 4.8+-1.96*2.1 (4.8+-4.2)Ex. 20a) For a flight with 400 seats, the airline can be 95% confident that between 87.06% and92.94% percent of the people will actually show up.p=.9[pic 1] where p = 0.90, [pic 2]= 1.959963985, n = 400[pic 3]= (0.8706, 0.9294)b) The approximate number of tickets should they sell so that there is less than a 10% chance they run out of seats can be calculated by trial and error method.We know that there 400 seats in the flight and only about 90% of the people who but the ticket actually show up.Let X be the number of tickets actually show up. Clearly X is binomial with n = 400 and p = 0.90.  The probability mass function binomial variable is given by[pic 4].