(16.10) An SRS of 350 high school seniors gained an average of xx¯ = 22 points i
ID: 3220336 • Letter: #
Question
(16.10) An SRS of 350 high school seniors gained an average of xx¯ = 22 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation 53.
Find a 99% confidence interval for based on this sample.
a) Confidence interval (±±0.01) is between ___ and ___
b) What is the margin of error (±±0.01) for 99%? ____
c) Suppose we had an SRS of just 100 high school seniors. What would be the margin of error (±±0.01) for 99% confidence? ____
(16.10) An SRS of 350 high school seniors gained an average of xx¯ = 22 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation 53.
Find a 99% confidence interval for based on this sample.
a) Confidence interval (±±0.01) is between ___ and ___
b) What is the margin of error (±±0.01) for 99%? ____
c) Suppose we had an SRS of just 100 high school seniors. What would be the margin of error (±±0.01) for 99% confidence? ____
Explanation / Answer
(a)
n = 350
x-bar = 22
s = 53
% = 99
Standard Error, SE = /n = 53 /350 = 2.832969164
z- score = 2.575829304
Width of the confidence interval = z * SE = 2.57582930354892 * 2.8329691642717 = 7.297244989
Lower Limit of the confidence interval = x-bar - width = 22 - 7.29724498938152 = 14.70275501
Upper Limit of the confidence interval = x-bar + width = 22 + 7.29724498938152 = 29.29724499
The 99% confidence interval is [14.70, 29.30]
(b) Margin of error = 7.30
(c)
n = 100
x-bar = 22
s = 53
% = 99
Standard Error, SE = /n = 53 /100 = 5.3
z- score = 2.575829304
Width of the confidence interval = z * SE = 2.57582930354892 * 5.3 = 13.65189531
Margin of error = 13.65
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