Confidence Intervals An SRS of 400 high school seniors gained an average of x =

Question

Confidence Intervals

An SRS of 400 high school seniors gained an average of x = 12 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation sigma = 42. We want to estimate the mean change in score mu in the population of all high school seniors. (a) Give a 95% confidence interval for mu based on this sample. (b) Based on your confidence interval in part (a), how certain are you that the mean change in score mu in the population of all high school seniors is greater than 0?

Explanation / Answer

a.
Confidence Interval
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)=12
Standard deviation( sd )=42
Sample Size(n)=400
Confidence Interval = [ 12 ± Z a/2 ( 42/ Sqrt ( 400) ) ]
= [ 12 - 1.96 * (2.1) , 12 + 1.96 * (2.1) ]
= [ 7.88,16.12 ]

b.
no, it does n't include the mean value to zero

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