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Tpy6a [65]
3 years ago
12

A field test for a new exam was given to randomly selected seniors. The exams were graded, and the sample mean and sample standa

rd deviation were calculated. Based on the results, the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 5% of 75%.
Is the confidence interval at 90%, 95%, or 99%? What is the margin of error? Calculate the confidence interval and explain what it means in terms of the situation.
Mathematics
1 answer:
IgorC [24]3 years ago
4 0
Hello there
<span>
A field test for a new exam was given to randomly selected seniors. The exams were graded, and the sample mean and sample standard deviation were calculated. Based on the results, the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 5% of 75%.

Is the confidence interval at 90%, 95%, or 99%? What is the margin of error? Calculate the confidence interval and explain what it means in terms of the situation.
</span>
4.8%
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Read 2 more answers
A^2(64a^2-3)/3=27-4a^2/4
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The answer is a = 3/4 = 0.75

First get rid of the paranthesis,
{a}^{2} (64 {a}^{2}  - 3) = 64 {a}^{4} -  3 {a}^{2}
Then set the denominators equal:
\frac{ 256 {a}^{4}  - 12 {a}^{2}}{12}  =  \frac{81 - 12 {a}^{2} }{12}
Then remove the denominators and solve:
256 {a}^{4}  - 12 {a}^{2}  = 81 - 12 {a}^{2}
Eliminate -12a^2 by adding 12a^2 to both sides:
256 {a}^{4}  = 81
Take the fourth root of them or take the square root twice:
\sqrt[4]{256 {a}^{4} }  =  \sqrt[4]{81}   \\  4a = 3
Divide both sides by 4:
a =  \frac{3}{4}  = 0.75
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