Question

When determining the sample size required to estimate , we always round the value of n up to the next whole number. Explain why we do so. (b) An estimate is needed of the mean acreage of farms in Ontario, Canada. A 95% confidence interval should have a margin of error of 25 acres. A study ten years ago in this province had a sample standard deviation of 200 acres for farm size. About how large a sample of farms is needed? (c) After randomly sampling 100 students in the student union to find out how many credit hours they were taking, a researcher constructed a confidence interval of (13.905, 15.095). Given that s = 3, what confidence level was the researcher using?

Answer 1

(a) We round the value of n, as a sample size cannot be in fractions (or decimals), irrespective whether the sample are living or non-living. Samples will have to take on positive, integer values.

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(b) Given ME = 25 acres, = 200 acres, = 0.05

The Zcritical at = 0.05 is 1.96

The ME is given by :

ME = Zcritical * \frac{\sigma}{\sqrt{n}}

Squaring both sides we get: (ME)² = (Z critical)² * ²/n

Therefore n = (Zcritical * /ME)² = (1.96*200/25)² = 245.8624

Therefore n = 246 (Taking it to the next whole number)

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(c) The Confidence interval is given by ME

ME = tc * Standard error, where the standard error = s/Sqrt(n)

We use tc values as population standard deviation is unknown.

Here given that s = 3, n = 100, therefore square root of 100 = 10.

Therefore Standard error = 3/10 = 0.3

Therefore ME = 0.3 * tc

The Upper Limit : 0.3 * tc = 15.095

The Lower Limit : 0.3 * tc = 13.905

Subtracting the 2 equations, we get 0.3 * tc - ( 0.3 * tc) = 15.095 - 13.905

Therefore 2 * 0.3 * tc = 1.19

tc = 1.9833

The probability at 1.9833 degrees of freedom = 99 is 0.025 (from t tables for 1 tail)

Therefore for 2 tails = 2 * 0.025 = 0.05

Therefore the confidence level being used is 100 - 5 = 95%

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