A private opinion poll is conducted for a politician favors decriminalizing marijuana possession. How large a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by more than 2%.

Hmm, let me put on my thinking clown shoes and calculate that for you.

Okay, so first we need to know the estimated proportion of people who favor decriminalizing marijuana possession in the entire population. Let's assume that's p = 0.5 (because hey, clowns like to be balanced).

The formula to calculate the sample size required is:

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 (95% in this case), and E is the maximum allowable error (2% in this case).

Now, let's plug in some numbers and make some magic happen. Using a standard normal distribution table, the corresponding z-score for a 95% confidence level is approximately 1.96.

So, the equation becomes:

n = (1.96^2 * 0.5 * (1-0.5))/(0.02^2)

Calculating this, we get:

n ≈ 384.16

But, since we can't have a fraction of a person in our sample, we'll round it up to the next whole number. So, you would need a sample size of at least 385 people to be 95% confident that the sample proportion will not differ from the true proportion by more than 2%.

Remember, though, this is just a rough estimate based on some assumptions. The real world is a lot messier. So take it with a grain of confetti!

To determine the sample size needed for a private opinion poll with a confidence level of 95% and a margin of error of 2%, we can use the formula for sample size for proportion:

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

Where:
n = sample size
Z = Z-value corresponding to the desired confidence level
p = estimated proportion of the population in favor of decriminalizing marijuana possession
E = margin of error (expressed as a proportion)

Since the true proportion is unknown, a common approach is to assume a conservative estimate of p (0.5), which provides the maximum sample size. This is done to ensure the sample size is not underestimated.

Using a Z-value of 1.96 for a 95% confidence level and a margin of error (E) of 2% (0.02), the formula becomes:

n = (1.96^2 * 0.5 * (1-0.5)) / 0.02^2

Simplifying:
n = (3.8416 * 0.25) / 0.0004
n = 9604

Therefore, a sample size of approximately 9604 respondents is needed to be 95% confident that the sample proportion will not differ from the true proportion by more than 2%.

To determine the sample size needed for a given confidence level and margin of error in estimating a proportion, we can use the formula:

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

where:
n = required sample size
Z = Z-score corresponding to the desired confidence level (for 95% confidence level, Z ≈ 1.96)
p = estimated proportion (assumed to be 0.5 when unknown, which maximizes the sample size)
E = desired margin of error (in this case, 0.02)

Now let's calculate the sample size using these values:

n = (1.96^2 * 0.5 * (1-0.5)) / 0.02^2

Simplifying the equation:

n = (3.8416 * 0.25) / 0.0004
n = 0.9604 / 0.0004
n = 2401

Therefore, a sample size of 2401 is needed to be 95% confident that the sample proportion will not differ from the true proportion by more than 2%.

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 95% confidence, p = .5 (when no value is stated in the problem), q = 1 - p, ^2 means squared, * means to multiply, and E = .02.

Plug values into the formula and calculate n.

I hope this will help get you started.