Linda Silbers is a social scientist studying voter opinion about a city bond proposal for a light rail mass transit system in Denver
If Linda wants to be 90 percent Confident that her sample estimate of the proportion of voters who favor the bond is within 2 percent of the population percent p who favor the bond how large a sample should she use?
n=(z*/m)squared(p*)(1-p*)
so be sure to square (z*/m)
m= margin of error, the invNorm for 90% is 1.645, p*=.5, if not given
The problem worked out:
(1.645/0.02)squared(.5)(1-.5)= 1691.265625 so the sample size should be 1691 people. Hope this helps
To determine how large a sample Linda should use to be 90 percent confident that her sample estimate of the proportion of voters who favor the bond is within 2 percent of the population percent p, she can use the formula for sample size calculation:
n = (Z^2 * p * (1 - p)) / (E^2)
where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, 90 percent confidence level can be represented by a Z-score of 1.645)
p = estimated proportion of voters who favor the bond (Linda can use a conservative estimate, say 0.5, if she doesn't have any prior information)
E = margin of error, which is 2 percent or 0.02
Plugging the values into the formula, we get:
n = (1.645^2 * 0.5 * (1 - 0.5)) / (0.02^2)
Simplifying further:
n = (2.706025 * 0.25) / 0.0004
n = 0.67650625 / 0.0004
n ā 1691.26
Therefore, Linda should use a sample size of approximately 1692 voters to be 90 percent confident that her sample estimate is within 2 percent of the population proportion.
To determine the sample size Linda should use, we need to consider the desired level of confidence, the margin of error, and the population proportion.
1. Confidence Level: Linda wants to be 90% confident in her results. This means that there is a 90% probability that the true population proportion lies within the margin of error of her sample estimate.
2. Margin of Error: Linda wants her sample estimate to be within 2 percent of the population proportion. In other words, she wants to be within 0.02 of the true population proportion.
To calculate the sample size, we can use the formula:
n = (Z^2 * p * (1-p)) / E^2
where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, for 90% confidence level, Z ā 1.645)
p = estimated population proportion (we don't know this yet)
E = margin of error
Since p is unknown, we can use 0.5 as a conservative estimate. This will give us the largest sample size needed (since p = 0.5 results in the highest variability).
Plugging in the values into the formula, we get:
n = (1.645^2 * 0.5 * (1-0.5)) / 0.02^2
n = (2.706 * 0.5 * 0.5) / 0.0004
n = 0.6765 / 0.0004
n ā 1691.25
Therefore, Linda should use a sample size of approximately 1692 voters to ensure a 90% confidence level with a margin of error of 2 percent.