Question

1. A sample survey of 54 discount brokers showed that the mean price charged for a trade of 100 shares at $50 per share was S33.77 and a standard deviation of S15. Develop a 95% confidence interval for the mean price charged by discount brokers for a trade of 100 shares at $50 per share. a) 6 marks b) Explain what the interval you found tells you 2 marks e) What sample size would be necessary to achieve a margin of error of t S2? Usea confidence level of 95%. 6 marks

Answer 1

mean = 33.77 and sd = 15

a)
z-value for 95% CI = 1.96
SE = std.dev./sqrt(n)  
= 15/sqrt(54)
= 2.0412

CI = (33.77 - 1.96*2.0412, 33.77 + 1.96*2.0412)
= (29.7692 , 37.7708 )

b)
The above interval explains , if a sample of 54 discount brokers is selected, one can be 95% confident that the mean price charged for a trade lies between (29.7692 , 37.7708 )

c)
Given CI Level is 95%
Margin of Error(ME)   = 2
std. dev. = 15
z-value of 95% CI = 1.9600
n = (z*sigma/ME)^2
n = (1.96*15/2)^2
= 216

d)
As the price $39.25 does not lie in the above CI, we can conclude that the price has dropped significantly