A simple random sample of 20 observations is derived from a normally distributed population with a known standard deviation of 3.2. 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 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.

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ogorwyne ogorwyne · Answer 1 · 2020-08-13T03:37:42+02:00

Answer:

1. It is satisfied

2. 1.4

3. 1.18

4. 95% confidence is wider

Explanation:

1. It is normally distributed since n<30

2. Margin of error with 95% confidence

= Alpha = 1 - 0.95

= O.05

Alpha/2 = 0.025

Z(0.025) = 1.960

Margin of error = z(1.960)*SD/√n

= 1.960*(3.2/√20)

= 1.960 x 0.7156

= 1.4025

Approximately 1.4

3. At 90%

Alpha = 1 -0.9

= 0.10

Alpha/2 = 0.05

Z(0.05) =1.645

E = 1.645 x 3.2/√20

= 1.645 x 0.7176

= 1.177

Approximately 1.18

4. From the calculations in 2 and 3 it is obvious that the margin of error with 95% confidence interval is wider.