SOLUTION: 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 revea

Algebra ->  Probability-and-statistics -> SOLUTION: 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 revea      Log On


   

Question 1208817: 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.)
Found 2 solutions by CPhill, Boreal:
Answer by CPhill(1987) About Me  (Show Source):
You can put this solution on YOUR website!
Here's how to calculate the necessary sample size:
**1. Identify Key Values:**
* **Confidence Level:** 99%
* **Margin of Error (E):** 2% = 0.02
* **Pilot Survey:** 8 out of 47 have two or more jobs. This gives us a preliminary estimate of the proportion (p̂): p̂ = 8/47 ≈ 0.1702
**2. Find the Z-score:**
For a 99% confidence level, the alpha (α) is 1 - 0.99 = 0.01. Since it's a two-tailed test, we divide alpha by 2: 0.01 / 2 = 0.005. We want the z-score that corresponds to an area of 0.995 (1 - 0.005) in the standard normal distribution table.
Z-score ≈ 2.58 (rounded to two decimal places).
**3. Use the Sample Size Formula:**
Since we have a preliminary estimate of the proportion from the pilot study, we use the following formula for sample size:
n = (Z² * p̂ * (1 - p̂)) / E²
Where:
* n = sample size
* Z = Z-score
* p̂ = estimated proportion
* E = margin of error
**4. Plug in the Values and Calculate:**
n = (2.58² * 0.1702 * (1 - 0.1702)) / 0.02²
n = (6.6564 * 0.1702 * 0.8298) / 0.0004
n = 0.9457 / 0.0004
n ≈ 2364.25
**5. Round Up:**
Always round the sample size *up* to the next whole number to ensure the desired confidence level and margin of error are met.
n = 2365
**Answer:**
You should interview 2365 people in the workforce to meet your requirements.

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
the point estimate is 0.170 and the sd=sqrt (0.170*0.830)/47)=0.053); that *2.58 z(0.995) =0.14 margin of error.
99% CI for the estimate is (0.03, 0.31)
margin of error=z*sqrt(0.17*0.83)/n
.02=2.58*sqrt(.17*.83)/n)
square both sides and rearrange
0.0004n=6.66*0.14
n=2331