Question 1208818: You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 99% confidence level and a margin of error of 2%. A pilot survey reveals that 8 of the 47 sampled hold two or more jobs. (Use t Distribution Table & z Distribution Table.)
How many in the workforce should be interviewed to meet your requirements? (Round z-score to 2 decimal places. Round up your answer to the next whole number.)
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **1. Determine the Sample Proportion from the Pilot Study**
* Calculate the sample proportion (p̂) of workers with two or more jobs:
p̂ = (Number of workers with two or more jobs) / (Sample size)
p̂ = 8 / 47
p̂ ≈ 0.1702
**2. Determine the Critical Value (zα/2)**
* For a 99% confidence level, the significance level (α) is 1 - 0.99 = 0.01.
* Since we're dealing with a two-tailed test, we need to find zα/2, where α/2 = 0.01 / 2 = 0.005.
* Using a standard normal (z) distribution table, find the z-score that corresponds to a cumulative probability of 0.995 (1 - 0.005).
* zα/2 ≈ 2.58
**3. Determine the Margin of Error (E)**
* The desired margin of error is given as 2% or 0.02.
**4. Calculate the Required Sample Size (n)**
* Use the following formula to calculate the required sample size:
n = (zα/2)² * p̂ * (1 - p̂) / E²
* Substitute the values:
n = (2.58)² * 0.1702 * (1 - 0.1702) / 0.02²
n = 6.6564 * 0.1702 * 0.8298 / 0.0004
n ≈ 248.48
* **Round up to the next whole number:**
n = 249
**Therefore, you would need to interview at least 249 individuals in the workforce to meet your requirements of a 99% confidence level and a margin of error of 2%.**
**Note:**
* This calculation assumes that the pilot study provides a reasonable estimate of the true population proportion.
* In practice, it's always a good idea to slightly increase the sample size to account for potential non-response or other unforeseen factors.
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