Question

How many computers?
In a simple random sample of 185 households, the sample mean number of personal computer ls was 1.64. assume the population of the standard deviation is 0.47
a) construct a 99.9% confidence interval for the mean number of personal computers. round the answer to at least two decimal places.
a 99.9% confidence interval for the mean number of personal computers is ---< mean <-----

2) In a sample of 75 clamps, the mean step to complete this step was 55.4 seconds. Assume that the population standard deviation is standard deviation 12. round the critical value to no less than 3 decimal places.
a) construct a 99% confidence interval for the mean time needed to complete this step. round your answer to at least one decimal place.
a 99% confidence interval for the mean is ----< mean <---------

Answer 1

Therefore, Therefore, Given = n = 75 Given = n = 185 o= 0.47 u = 1.64 Standard error = c/Vn = 0.47/7185 = 0.0345551 Standard error = c/Vn = 12/175 = 1.3856406 Lower limit = o = 12 Lower limit = u = 55.4 At 99.9 % confidence level = = X-E = 1.64-0.114 = 1.526 At 99 % confidence level = = X-E = 55.4-3.569 = 51.831 Now, Now, a = 1 -99.9% a = 1 -99% Margin of error (E) = Upper limit = Margin of error (E) = Upper limit = = 1 - 0.999 = 1E-03 Therefore, a/2 = 5E-04 = 1 -0.99 = 0.01 Therefore, a/2 = 0.005 = X + E = Standard error * Za|2 = 0.0346*3.2905 = 0.1137 = X +E = 1.64+0.114 = 1.754 nu = Standard error * Zal2 = 1.3856*2.5758 = 3.5692 = 55.4+3.569 = 58.969 Za /2 = 3.291 Za /2 = 2.576

Pls upvote...!!!