The percentage of people not covered by health care insurance in 2003 was 15.6% (Statistical Abstract of the United States, 2006). A congressional committee has been charged with conducting a sample survey to obtain more current information.
a. What sample size would you recommend if the committee's goal is to estimate the current proportion of individuals without health care insurance with a margin of error of .03 (to the nearest whole number)? Use a 95% confidence level.
b. Repeat part (a) using a 99% confidence level (to the nearest whole number).
a. Hmm, let me crunch some numbers for you. To estimate the current proportion of individuals without health care insurance with a margin of error of 0.03 at a 95% confidence level, you'll need a sample size of approximately 1,068 people. Now, that's quite a party!
b. Feeling extra confident, huh? For a 99% confidence level, I'd recommend a larger sample size. You would need around 1,815 people to achieve that margin of error of 0.03. With that many people, it'll be like hosting a clown convention!
a. To find the sample size needed with a 95% confidence level and a margin of error of 0.03, 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 (95% => z = 1.96)
p = estimated proportion of individuals without health care insurance (0.156)
E = margin of error (0.03)
Plugging in the values, we get:
n = (1.96^2 * 0.156 * (1-0.156)) / 0.03^2
n = (3.8416 * 0.156 * 0.844) / 0.0009
n = 0.5049 / 0.0009
n ≈ 560.99
Rounded to the nearest whole number, the recommended sample size is 561.
b. To find the sample size needed with a 99% confidence level and a margin of error of 0.03, we use the same formula as in part (a), but with a different z-score (99% => z = 2.576):
n = (2.576^2 * 0.156 * (1-0.156)) / 0.03^2
n = (6.6518 * 0.156 * 0.844) / 0.0009
n = 0.8844 / 0.0009
n ≈ 982.67
Rounded to the nearest whole number, the recommended sample size is 983.
To calculate the sample size needed to estimate the proportion of individuals without health care insurance, 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)
p = estimated proportion (15.6% or 0.156 in this case)
E = margin of error
a. For a 95% confidence level:
The Z-score for a 95% confidence level is approximately 1.96.
Using the given information, we plug these values into the formula:
n = (1.96^2 * 0.156 * (1 - 0.156)) / 0.03^2
n = 1624.97
Rounded up to the nearest whole number, the sample size needed is approximately 1625.
Therefore, I would recommend a sample size of 1625 for a 95% confidence level.
b. For a 99% confidence level:
The Z-score for a 99% confidence level is approximately 2.576.
Using the given information, we again plug the values into the formula:
n = (2.576^2 * 0.156 * (1 - 0.156)) / 0.03^2
n = 4274.97
Rounded up to the nearest whole number, the sample size needed is approximately 4275.
Therefore, I would recommend a sample size of 4275 for a 99% confidence level.