A researcher at a major hospital wishes to estimate the proportion of the adult population of the U.S. that has high blood pressure. How large a sample is needed in order to be 99% confident that the sample proportion will not differ from the tgrue proportion by more than 4%?
Formula:
n = [(z-value^2 * p * q)]/E^2
...where n = sample size, z-value will be 2.58 using a z-table to represent the 99% confidence interval, p and q represent proportions, E = .04, ^2 means squared, and * means to multiply.
Note: p and q will both be .5 if no value is stated.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
Why did the blood pressure researcher become a comedian? Because they realized it's all about the "punny" percentages! Now, let's calculate the sample size they need with a dose of humor!
To estimate the required sample size, we need to consider the level of confidence, allowable margin of error, and the estimated proportion. In this case, the researcher wants to be 99% confident and have a margin of error of 4%.
Now, let me juggle the statistical formula for you:
n = (Z^2 * p * (1 - p)) / (E^2)
Where:
n = sample size needed
Z = Z-score for desired level of confidence (99% confidence corresponds to a Z-score of approximately 2.576)
p = estimated proportion (let's assume 0.5, which gives the maximum sample size to be on the safe side)
E = allowable margin of error (4% or 0.04)
Substituting these values into the equation, we have:
n = (2.576^2 * 0.5 * (1 - 0.5)) / (0.04^2)
Now, let me juggle these numbers...
The sample size needed is approximately 665.38. However, we can't have a fraction of a participant, so let's round it up to 666. Don't worry, no devilish statistics here!
Therefore, the researcher needs a sample size of 666 adults to be 99% confident that the sample proportion will not differ from the true proportion by more than 4%. Now that's a size fit for a circus!
To determine the sample size needed to estimate the proportion of the adult population of the U.S. with high blood pressure, you can use the following formula:
n = (Z^2 * p * (1 - p)) / E^2
Where:
n = sample size
Z = Z-score representing the desired level of confidence (in this case, 99% confidence, so Z = 2.58)
p = estimated proportion (if you don't have an estimate, you may assume p = 0.5 for maximum variability)
E = maximum allowable error (4% in this case, so E = 0.04)
Using these values, you can plug them into the formula:
n = (2.58^2 * 0.5 * (1 - 0.5)) / 0.04^2
Calculating this:
n = (6.6564 * 0.25) / 0.0016
n = 16.641 / 0.0016
n ≈ 10,400
Therefore, a sample size of approximately 10,400 is needed in order to be 99% confident that the sample proportion will not differ from the true proportion by more than 4%.
To determine the sample size needed to estimate the proportion of the adult population with high blood pressure, we can use the following formula:
n = (Z^2 * p * (1-p)) / E^2
Where:
n = required sample size
Z = Z-score corresponding to the level of confidence (in this case, 99% confidence level)
p = estimated proportion (we can assume 0.5 as a conservative estimate since we do not have an accurate estimate)
E = maximum tolerable error (4% = 0.04)
Now let's calculate the sample size using the formula:
Z = 2.58 (For a 99% confidence level, the Z-score is 2.58 based on a standard normal distribution)
p = 0.5 (conservative estimate)
E = 0.04
n = (2.58^2 * 0.5 * (1-0.5)) / 0.04^2
n = (6.6564 * 0.25) / 0.0016
n = 1.6641 / 0.0016
n ≈ 1038.81
Since you can't have a fraction of a person, you'll need to round up to the nearest whole number. Therefore, the researcher would need a sample size of approximately 1039 individuals to be 99% confident that the sample proportion will not differ from the true proportion by more than 4%.