Question

You measure the weight of 40 bags of nuts, and find they have a mean weight of 64 ounces. Assume the population standard deviation is 2.4 ounces. Based on this, what is the maximal margin of error associated with a 92% confidence interval for the true population mean bags of nuts weight.

Give your answer as a decimal, to two places

m =  ounces

Answer 1

Solution

Given that,

= 64

= 2.4

n = 40

At 92% confidence level the z is ,

  = 1 - 92% = 1 - 0.92 = 0.08

/ 2 = 0.08 / 2 = 0.04

Z/2 = Z0.04 = 1.751

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

= 17.51 * (2.4 / 40 )

= 0.66

Margin of error = 0.66