SOLUTION: A university system enrolling hundreds of thousands of students is considering a change in the way students pay for their education. Currently, the students pay $400 per credit ho

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Question 1082018: A university system enrolling hundreds of thousands of students is considering a change in the way students pay for their education. Currently, the students pay $400 per credit hour. The university system administrators are contemplating charging each student a set fee of $7,000 per quarter, regardless of how many credit hours each takes. To see if this proposal would be economically feasible, the administrators would like to know how many credit hours, on the average, each student takes per quarter. A random sample of 250 students yields a mean of 14.1 credit hours per quarter and a standard deviation of 2.3 credit hours per quarter. Suppose the administration wanted to estimate the mean to within 0.1 hours at 95% reliability and assumed that the sample standard deviation provided a good estimate for the population standard deviation. How large a total sample would they need to take?

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The formula we'll use is
n+=+%28+%28z%2As%29%2FE+%29%5E2

where
n = minimum sample size
z = standard normal critical value
s = sample standard deviation, which approximates the population standard deviation sigma (sigma)
E = margin of error

We're given
s = 2.3
E = 0.1

At 95% confidence, the critical value is
z = 1.96
which is approximate.
This value is found using a calculator or a table.

Let's plug s = 2.3, E = 0.1 and z = 1.96 into the formula to get

n+=+%28+%28z%2As%29%2FE+%29%5E2

n+=+%28+%281.96%2A2.3%29%2F0.1+%29%5E2

n+=+%28+%284.508%29%2F0.1+%29%5E2

n+=+%28+45.08+%29%5E2

n+=+2032.2064

n+=+2033 Round up to the nearest whole number

We round up to ensure we clear the hurdle. If n = 2032 then the sample size isn't large enough to make E = 0.1

Note how plugging n = 2032 into the margin of error formula E+=+z%2A%28s%2Fsqrt%28n%29%29 yields
E+=+z%2A%28s%2Fsqrt%28n%29%29
E+=+1.96%2A%282.3%2Fsqrt%282032%29%29
E+=+0.1000050786112
The margin of error (E) is larger than 0.1 so this sample size isn't large enough.

While plugging n = 2033 into the margin of error formula yields
E+=+z%2A%28s%2Fsqrt%28n%29%29
E+=+1.96%2A%282.3%2Fsqrt%282033%29%29
E+=+0.09998048014111
telling us we have cleared the threshold needed. We now have E < 0.1 making this the smallest sample size possible. Anything larger for n and E simply gets smaller.

note: "margin of error within 0.1" means that the error is either less than 0.1 or at 0.1 exactly.
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Answer: The minimum sample size needed is 2033