Suppose you had been in charge of designing the study. what sample size would be needed to construct a margin of error of 2% with 95% confidence? use the prior point estimate of p* = 0.15 for this calculation. round up to the nearest whole number. (for example, 144.1 would round to 145)

1225

Step-by-step explanation:

hihi. So the equation for MoE is (z*) * SE. The z* for a 95% Confidence is one you should have memorized but for repeatability sake you can always just do an inverse Norm to find the z* for these types of applications. To do so, you can always type this command into your calculator: invNorm(conf + (1-conf)/2, 0, 1).

(When I say conf here I am referring to the confidence level as a decimal).

All that's left is the Standard Error or SE to be short. Since you gave a p* estimate then we can use the equation for SE when dealing with proportions/percents which is sqrt(p(1-p) / n) where p is the proportion and n is the sample size, which we are solving for. Once you have this established it's a basic multi-step solve for n which comes out to be 1225 after rounding.

A side note, the included picture is a bit messy due to my refusal to round when doing these kinds of problems. Rounding errors are more common than you think

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632