Question

To estimate the mean score ? of those who took the Medical College Admission Test on your campus, you will obtain the scores of a random sample of students. From published information, you know that the scores are approximately Normal with standard deviation about 6.5. You want your sample mean (x ?) to estimate ? with an error of no more than 1 point in either direction. 1)What standard deviation must x ? have so that 99.7% of all samples give an x ? within 1 point of ?. 2) How large a random sample do you need in order to reduce the standard deviation of x ? to the value you found in part 1?

Answer #1

Here as 99..7% of all samples must have sample mean withing 1 point of sample mean.

so margin of error = 1

Population standard deviation = 6.5

margin of error = critical test statistic * standard error of sample mean

critical test statistic = 3(as 99.7% of samples will lie in between these values)

standard deviation of = 1/3 = 0.3333

(2) Here if the required sample size n = n

standard error of sample mean = 6.5/sqrt(n)

6.5/sqrt(n) = 0.3333

sqrt(n) = 6.5 * 3

n = 380.25

n = 381

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to take the Medical College Admission Test (MCAT). To estimate the
mean score ?μ of those who took the MCAT on your campus, you will
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to take the Medical College Admission Test (MCAT). To estimate the
mean score ? of those who took the MCAT on your campus, you will
obtain the scores of an SRS of students. The scores follow a Normal
distribution, and from published information you know that the
standard deviation is 10.4 . Suppose that, unknown to you, the mean
score of those taking the MCAT on your campus is 510...

Almost all medical schools in the United States require students
to take the Medical College Admission Test (MCAT). To estimate the
mean score ?μ of those who took the MCAT on your campus, you will
obtain the scores of an SRS of students. The scores follow a Normal
distribution, and from published information you know that the
standard deviation is 10.8 . Suppose that, unknown to you, the mean
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to take the Medical College Admission Test (MCAT). To estimate the
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distribution, and from published information you know that the
standard deviation is 10.8 . Suppose that, unknown to you, the mean
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