Question

A simple random sample of 32 observations is derived from a normally distributed population with a known standard deviation of 2.1. [You may find it useful to reference the z table.] a. Is the condition that X− is normally distributed satisfied? Yes No b. Compute the margin of error with 95% confidence. (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 with 90% confidence. (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 interval? The margin of error with 95% confidence. The margin of error with 90% confidence.

Answer 1

We have given n = 32 , population standard deviation = 2.1

a) We know that if X follows Normal distribution then the sample mean taken from the X is also follows Normal distribution.

Yes, condition that X− is normally distributed satisfied.

b) Margin of error with C= 95%

We know that Margin of error E = Z_c * ( / n)

For 95% we get Margin of error E = Z0.95 * ( 2.1 / 32)

= 1.96 * ( 2.1 / 32)

= 0.73

For 95% we get Margin of error E = 0.73

c) Margin of error with C= 90%

For 90% we get Margin of error E = Z_0.90 * ( 2.1 / 32)

= 1.645 * ( 2.1 / 32)

= 0.61

For 90% we get Margin of error E = 0.61

d) From b) and c) we can say that The margin of error with 95% confidence will lead to a wider interval.

Hope this will help you.Thank you :)