An SRS of 500 high school seniors gained an average of x 21

An SRS of 500 high school seniors gained an average of x = 21 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation 52 .

Find a 90 % confidence interval for based on this sample.

Confidence interval (±0.01) is between and

What is the margin of error (±0.01) for 90 %?

Suppose we had an SRS of just 100 high school seniors. What would be the margin of error (±0.01) for 90 % confidence?

Solution

a)
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=21
Standard deviation( sd )=52
Sample Size(n)=500
Confidence Interval = [ 21 ± Z a/2 ( 52/ Sqrt ( 500) ) ]
= [ 21 - 1.64 * (2.326) , 21 + 1.64 * (2.326) ]
= [ 17.186,24.814 ]
b)
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=21
Standard deviation( sd )=52
Sample Size(n)=500
Margin of Error = Z a/2 * 52/ Sqrt ( 500)
= 1.64 * (2.326)
= 3.814
c)
Sample Size(n)=100
Margin of Error = Z a/2 * 52/ Sqrt ( 100)
= 1.64 * (5.2)
= 8.528