An SRS of 400 high school seniors gained an average of x 22

An SRS of 400 high school seniors gained an average of x¯ = 22 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation 49 .

Find a 95 % confidence interval for based on this sample.

Confidence interval (±0.01) is between  and

What is the margin of error (±0.01) for 95 %?

Suppose we had an SRS of just 100 high school seniors. What would be the margin of error (±0.01) for 95 % confidence?

Solution

a)
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=22
Standard deviation( sd )=49
Sample Size(n)=400
Confidence Interval = [ 22 ± t a/2 ( 49/ Sqrt ( 400) ) ]
= [ 22 - 1.966 * (2.45) , 22 + 1.966 * (2.45) ]
= [ 17.183,26.817 ]
b)
Margin of Error = t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
Mean(x)=22
Standard deviation( sd )=49
Sample Size(n)=400
Margin of Error = t a/2 * 49/ Sqrt ( 400)
= 1.966 * (2.45)
= 4.817
c)
Margin of Error = t a/2 * (sd/ Sqrt(n))
Sample Size(n)=100
Margin of Error = t a/2 * 49/ Sqrt ( 100)
= 1.984 * (4.9)
= 9.722