MyOpenMath

Let's say we want to calculate the 90% confidence interval for proportion. Based on previous research, a reasonable guess for the population proportion is 20%. The desired margin of error is 2%.

Detailed Solution

We have the desired value for $M E = 0.02$ and since a guess/estimate for the population proportion is known, we'll use an estimate for the proportion $p = 0.2$ . The only thing missing for the formula is the critical value. You can find the critical z-value, $z_{c}$ , for the 90% confidence level by using Desmos, Statkey, or a Ti83/84+ calculator. You may also use Excel by entering the following into an Excell cell:

=NORMSINV((1+90%)/2) which returns the value

z_{c} = 1.6448536

. Plugging these values in, we get

n = ⌈ {1.6448536}^{2} \cdot \frac{0.2 \cdot (1 - 0.2)}{{0.02}^{2}} ⌉ = ⌈ 1082.2173978 ⌉ = 1083

Again, note that we always round the sample size up to the nearest integer.

Thus, if we have a sample with a proportion close to $20 %$ , a sample of size $n = 1083$ will guarantee that the margin of error for a 90% confidence interval will be within $M E = 2 %$ of the true population proportion.

Practice Problem 1
If a resonable guess for the proportion is 34%, find the necessary sample size to find a 90% confidence level with a margin of error of 2%:

n =

Practice Problem 2
Now, if you find yourself in a situation where you do not have a reasonable guess for the population proportion, then you can always use the worst-case scenario. It turns out, the largest sample size will be needed when the population proportion is $50 %$ or $0.5$ or $\frac{1}{2}$ . So the formula for sample size becomes $n = 0.5 (1 - 0.5) {(\frac{z_{c}}{M E})}^{2} = 0.25 {(\frac{z_{c}}{M E})}^{2}$ .

Find the necessary sample size for a 90% confidence level with a margin of error of 2% if a reasonable guess for the population is not available.

n =

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