You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 90 percent confidence level and state that the estimated proportion must be within 2 percent of the population proportion. A pilot survey reveals that 7 of the 50 sampled hold two or more jobs.


How many in the workforce should be interviewed to meet your requirements?

Oh, I'm no mathematician, but I am a fan of surveys. Let's see, you want a 90 percent confidence level with an estimated proportion within 2 percent of the population proportion. Well, since the pilot survey revealed that 7 out of 50 people had two or more jobs, we can estimate the proportion as 7/50.

Now, to figure out how many people you should interview, we can use a nifty little formula. It goes like this:

n = (Z^2 * p * (1-p)) / E^2

Where:
- n is the sample size you're looking for
- Z is the z-score for the desired confidence level (for 90 percent, it's approximately 1.645)
- p is the estimated proportion from the pilot survey (7/50 in this case)
- E is the desired margin of error (2 percent, or 0.02)

Alright, let's plug in those numbers and see what we get:

n = (1.645^2 * (7/50) * (1 - 7/50)) / (0.02^2)

After some number crunching, it looks like you would need to interview approximately 369 people from the workforce to meet your requirements. Good luck with your survey! And remember, if you need any clown-related questions answered, I'm here for you.

To determine the sample size required to meet the given requirements, we can use the formula:

n = ((Z * σ) / E)^2

Where:
n = Required sample size
Z = Z-score corresponding to the desired confidence level
σ = Standard deviation (which is the square root of the population proportion multiplied by 1 minus the population proportion)
E = Maximum error (in this case, 2 percent)

In this case, the information provided does not directly give us the population proportion. However, we can estimate it using the pilot survey results. From the pilot survey, 7 out of 50 sampled held two or more jobs. Therefore, the estimated proportion is 7/50 or 0.14.

Now, we need to determine the Z-score corresponding to the desired confidence level of 90 percent. The Z-score for a 90 percent confidence level is approximately 1.645.

Next, we need to calculate the standard deviation (σ) using the estimated proportion from the pilot survey. The formula for standard deviation is:

σ = sqrt((p * (1-p)) / n)

where p is the estimated proportion and n is the sample size (which is unknown).

Since we are trying to find the required sample size, we can rearrange the formula as:

n = (p * (1-p) * (Z / E)^2) / (E^2 + (p * (1-p) * (Z / E)^2))

Plugging in the values:

n = (0.14 * (1-0.14) * (1.645^2)) / (0.02^2 + (0.14 * (1-0.14) * (1.645^2)))

Calculating this, we find that the required sample size is approximately 299.