Question

In: Statistics and Probability

Suppose a random sample of size 40 was drawn from a normally distributed population, with a...

Suppose a random sample of size 40 was drawn from a normally distributed population, with a known population standard deviation of σ=6.4.

a) What is the margin of error for a 95% confidence level?

Round your response to at least 3 decimal places.

   

b) What is the margin of error for a 90% confidence level?

Round your response to at least 3 decimal places.

Solutions

Expert Solution

Solution

Given that,

= 6.4

n = 40

a ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.960 * (6.4 / 40 )

Margin of error = 1.983

b ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645* (6.4 / 40 )

Margin of error = 1.664


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