Question

How to calculate margin of error for this question?

Starting salaries of 80 college graduates who have taken a
statistics course have a mean of $42,893. Suppose the distribution
of this population is approximately normal and has a standard
deviation of $10,748.

Use a 93% confidence level.

Can't get the answer right :(

Answer #1

Solution :

Given that,

Population standard deviation = = $10748

Sample size n =80

At 93% confidence level the z is ,

= 1 - 93% = 1 - 0.93= 0.07

/ 2 = 0.035

Z / 2 = Z0.035 = 1.81 ( Using z table )

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

= 1.81 * ( 10748 / 80 )

E= 2175.0099

Starting salaries of 110 college graduates who have taken a
statistics course have a mean of $44,836. Suppose the distribution
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Using a 75% confidence level, find both of the following:
(a) The margin of error:
(b) The confidence interval for the mean μ: ____ < μ <
_____

(1 point) Starting salaries of 135 college graduates who have
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find both of the following:
A. The margin of error:
B. Confidence interval: ,

Starting salaries of 140 college graduates who have taken a
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Starting salaries of 64 college graduates who have taken a
statistics course have a mean of $42,500 with a standard deviation
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decimal places.)

Starting salaries of 64 college graduates who have taken a
statistics course have a mean of $43,500 with a standard deviation
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commas or dollar signs in your answers. Round each bound to three
decimal places.)
Lower-bound:
Upper-bound:

Question 3 4 pts
A sample of salaries of 57 college graduates who took a
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