Determine the sample size necessary to estimate a population proportion to within .03 with 90% confidence assuming you have no knowledge of the approximate value of the sample proportion.
Try this formula:
n = [(z-value)^2 * p * q]/E^2
Note: n = sample size needed; .5 for p and .5 for q are used if no value is stated in the problem. E = maximum error, which is .03 in the problem. Z-value is found using a z-table to represent 90% interval. Round final result to next highest whole number.
Also: ^2 means squared and * means to multiply.
I hope this will help get you started.
n = [(z-value)^2 * p * q]/E^2
To determine the sample size necessary to estimate a population proportion within .03 with 90% confidence, you can use the formula:
n = (z^2 * p * (1 - p)) / E^2
where:
n is the sample size
z is the z-score corresponding to the desired confidence level (90% confidence level corresponds to a z-score of 1.645)
p is the estimated proportion (since we have no knowledge of the approximate value, we can use p = 0.5 to get the maximum sample size)
E is the desired margin of error (in this case, 0.03)
Plugging in the values into the formula:
n = (1.645^2 * 0.5 * (1 - 0.5)) / 0.03^2
n = (2.705^2 * 0.25) / 0.0009
n = 6.890 / 0.0009
n ≈ 7664
Therefore, the sample size necessary to estimate a population proportion to within .03 with 90% confidence, assuming no knowledge of the approximate value of the sample proportion, is approximately 7664.
To determine the sample size necessary to estimate a population proportion with a specified margin of error and confidence level, you can use the formula:
n = (Z^2 * p * (1 - p)) / E^2
Where:
n is the sample size
Z is the Z-score corresponding to the desired confidence level (for 90% confidence level, Z = 1.645)
p is the estimated proportion of the population (since we have no knowledge of the approximate value, we can assume p = 0.5 for maximum variability)
E is the margin of error (in this case, E = 0.03)
Substituting the values into the formula:
n = (1.645^2 * 0.5 * (1 - 0.5)) / 0.03^2
n = (2.705 * 0.25) / 0.0009
n = 6.7625 / 0.0009
n ≈ 7518.06
Rounding up to the nearest whole number, the sample size necessary to estimate the population proportion to within 0.03 with 90% confidence is approximately 7519.