An SRS of 450 high school seniors gained an average of x ¯ ¯ ¯ x¯ = 22 points in

Question

An SRS of 450 high school seniors gained an average of x ¯ ¯ ¯ 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 45.

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? __________________

PLEASE ANSWER THE QUESTION AS ASKED, PROVIDE FOUR ANSWER, ONE ANSWER TO EACH BLANK SPACE

Explanation / Answer

for std error =std deviation/(n)1/2 where n=450

=2.1213

for 95% CI. z=1.96

henceConfidence interval =xbar -/+z*std error =17.842 ; 26.158

margin of error =z*std error =4.158

for n=100

std error =4.5

hence margin of error =8.820

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