a study is undertaken to determine the proportion of industry executives who believe that worker's pay should be based on individual performance. How many executives should be interviewed if an estimate is desired at the 99% confidence level to be withing .06? to be within .03? to be within .02?
How many executives should be interviewed if an estimate is desired at the 99% confidence interval to within ±0.06?
To calculate the sample size needed for estimating proportions, we can use the formula:
n = (Z^2 * p * q) / E^2
Where:
n = sample size
Z = z-score corresponding to the desired confidence level (e.g., 99% confidence level -> Z = 2.576)
p = estimated proportion (0.5 is often used as it maximizes the sample size)
q = 1 - p
E = acceptable margin of error (difference between the estimate and the true population value)
Now, let's calculate the sample size for each margin of error:
For within 0.06:
E = 0.06
Z = 2.576 (for 99% confidence level)
p = 0.5
q = 1 - p = 0.5
n = (2.576^2 * 0.5 * 0.5) / 0.06^2
n ≈ 426.256
Therefore, approximately 427 industry executives should be interviewed if an estimate with a margin of error of 0.06 is desired at the 99% confidence level.
For within 0.03:
E = 0.03
Z = 2.576 (for 99% confidence level)
p = 0.5
q = 1 - p = 0.5
n = (2.576^2 * 0.5 * 0.5) / 0.03^2
n ≈ 1705.025
Therefore, approximately 1706 industry executives should be interviewed if an estimate with a margin of error of 0.03 is desired at the 99% confidence level.
For within 0.02:
E = 0.02
Z = 2.576 (for 99% confidence level)
p = 0.5
q = 1 - p = 0.5
n = (2.576^2 * 0.5 * 0.5) / 0.02^2
n ≈ 3844.06
Therefore, approximately 3845 industry executives should be interviewed if an estimate with a margin of error of 0.02 is desired at the 99% confidence level.
To determine the sample size for estimating the proportion of industry executives who believe that worker's pay should be based on individual performance, we need to use the formula for sample size calculation:
n = [(Z * σ) / E]²
Where:
- n is the required sample size
- Z is the Z-score corresponding to the desired confidence level
- σ is the estimated standard deviation (in this case, we assume p * (1 - p), where p is the proportion we want to estimate)
- E is the desired margin of error
Let's calculate the sample sizes for each desired margin of error:
1. For a margin of error of 0.06:
Z = Z-score at 99% confidence level = 2.576 (looked up from the Z-table)
E = 0.06
n = [(2.576 * σ) / 0.06]²
2. For a margin of error of 0.03:
Z = Z-score at 99% confidence level = 2.576 (lookup from the Z-table)
E = 0.03
n = [(2.576 * σ) / 0.03]²
3. For a margin of error of 0.02:
Z = Z-score at 99% confidence level = 2.576 (lookup from the Z-table)
E = 0.02
n = [(2.576 * σ) / 0.02]²
To calculate σ, we need to make an initial assumption about the proportion. Since we don't have any prior information, a conservative estimate would be p = 0.5. We can always adjust the sample size later if the true proportion is significantly different.
Finally, we can compute the approximate sample sizes using the formulas above.
Formula to find sample size:
n = [(z-value)^2 * p * q]/E^2
... where n = sample size, z-value is found using a z-table for 99% confidence, p = .5 (when no value is stated in the problem), q = 1 - p, ^2 means squared, * means to multiply, and E = .06, .03, .02.
Plug values into the formula and calculate n. You will have 3 different answers using the E values above.
I hope this will help get you started.