Question

In: Statistics and Probability

Consider a population with a known standard deviation of 18.6. In order to compute an interval...

Consider a population with a known standard deviation of 18.6. In order to compute an interval estimate for the population mean, a sample of 58 observations is drawn. [You may find it useful to reference the z table.]

a. Is the condition that X−X−  is normally distributed satisfied?

  • Yes

  • No



b. Compute the margin of error at a 99% confidence level. (Round intermediate calculations to at least 4 decimal places. Round "z" value to 3 decimal places and final answer to 2 decimal places.)

c. Compute the margin of error at a 99% confidence level based on a larger sample of 245 observations. (Round intermediate calculations to at least 4 decimal places. Round "z" value to 3 decimal places and final answer to 2 decimal places.)

d. Which of the two margins of error will lead to a wider confidence interval?

  • 99% confidence with n = 245.

  • 99% confidence with n = 58.

Solutions

Expert Solution

Solution :

Given that,

Population standard deviation = = 18.6

Sample size = n = 58

a) Yes

b)

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2 =0.01/ 2= 0.005

Z/2 = Z0.005 = 2.576

Z/2 = 2.576

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

= 2.576 * (18.6 / 58 )

= 6.29

c)

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2 =0.01/ 2= 0.005

Z/2 = Z0.005 = 2.576

Z/2 = 2.576

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

= 2.576 * (18.6 / 245 )

= 3.06

d)99% confidence with n = 58.


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