Question

PART I

Wiley Publications has determined that out of a sample of 8,578 of its publications for 2012, 360 of them had been pirated online in some form. What is the estimate of the population proportion? What is the standard error of this estimate?

	1) Estimate of proportion: 0.042, Standard error: 0.0022.
	2) Estimate of proportion: 0.042, Standard error: 0.0000.
	3) Estimate of proportion: 0.958, Standard error: 0.0000.
	4) Estimate of proportion: 0.958, Standard error: 0.0022.
	5) The true population proportion is needed to calculate this.

PART II

In a recent survey of 123 WMU graduates, 89 students said that parking was too limited on campus. What is the estimate of the population proportion? What is the standard error of this estimate?

	1) Estimate of proportion: 0.276, Standard error: 0.0403.
	2) Estimate of proportion: 0.724, Standard error: 0.0403.
	3) Estimate of proportion: 0.724, Standard error: 0.0036.
	4) Estimate of proportion: 0.276, Standard error: 0.0036.
	5) The true population proportion is needed to calculate this.

PART III

A sample proportion is calculated from a sample size of 196. How large of a sample would we need in order to decrease the standard error by a factor of 8?

	1) 554
	2) 6,076
	3) 3,332
	4) 12,544
	5) 1,568

Answer 1

Part 1:

The point estimate of the proportion here is computed as:

p = x/n = 360/8578 = 0.042

The standard error here is computed here as:

Therefore 1) is the correct metrics here.

Part II)

Again, the point estimate of proportion here is computed as:
p = x/n = 89/123 = 0.724

Now the standard error here is computed as:

Therefore 2) is the correct answer here.

Part III)

Note that the standard error is inversely proportional to the square root of the sample size. Therefore to decrease the standard error by a factor of 8, the sample size need to be increased by a factor of 8² that is the new sample size should be computed as:

= 64*196 = 12544

Therefore 4) 12544 is the required sample size here.