A programmer plans to develop a new software system. In planning for the operati

Question

A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimate the percentage of computers that use a new operating system. How many computers must be surveyed in order to be 95% confident that his estimate is in error by no more than fourfour percentage points? Complete parts (a) through (c) below.

a. Assume that nothing is known about the percentage of computers with new operating systems. n=?

b. Assume that a recent survey suggests that about 99% of computers use a new operating system. n=?

c.  Does the additional survey information from part (b) have much of an effect on the sample size that is required?

Explanation / Answer

The assumption is made that we are interested in estimating a proportion of a normally distributed population.

Standard error is calculated by the formula z*square-root( p-hat *(1 - p-hat)/N)

where z is the z-score of the desired confidence interval, in this case, z=1.96.

N is the sample size that we are trying to calculate and p-hat is the sample proportion, which in this case is treated as a single degree of freedom (a free variable), typically some benchmark value.

Since we want the error to be within two percentage points, the equation becomes:

0.02 = 1.95*square-root( p-hat * ( 1 - p-hat) / N)

Solving for N in terms of p-hat, the equation becomes:

N = 9506.25 * p-hat * ( 1- p-hat)

So for example, using a benchmark sample proportion of p-hat = 0.5 (meaning half of the computers in the sample have the new OS), then N = 2377 computers will have to be sampled to ensure that the sample proportion is within 2% of the actual population proportion.

Moreover, differentiating N ( a parabolic function ) with respect to p-hat gives a first derivative of: 9506.25*(1 - 2*p-hat).

The maximum then is where this first derivative is zero.

So when 9506.25*( 1 - 2 *p-hat) = 0, then the maximum benchmark occurs at p-hat = 1/2 as verified in the table below.

The table summarizes the required sample sizes needed for various benchmark values.

For example, assuming the population percentage is 10% then only 856 computers need to be sampled.

0 0

0.1 855.5625

0.2 1521
0.3 1996.3125
0.4 2281.5
0.5 2376.5625 <-- average benchmark
0.6 2281.5
0.7 1996.3125
0.8 1521
0.9 855.5625
1 0

So, the answer is 2377 computers must be sampled

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