Question

You measure 27 backpacks' weights, and find they have an average weight of 57 grams. Assume the population standard deviation is 5 grams. Based on this, what is the margin of error associated with a 99% confidence interval for the true population mean backpack weight. The sample size is sufficiently large that the normal distribution value can be used in deriving the margin of error.

Answer 1

Solution :

Given that,

Population standard deviation =    = 5
Sample size = n =27

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

Margin of error = E = Z/2* ( /n)

=2.576 * (5 / 27)

E= 2.4788