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You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ (i.e., $K$ represents the number of colors, answers etc).

Let $p_k$ be the true proportion of people in the population who would choose the $k^\text{th}$ choice when presented with $K$ choices. You can re-interpret $p_k$ as the probability that a random person would choose the $k^\text{th}$ choice when presented with $K$ choices. Thus, by definition, we have:

$$\sum_{k=1}^K p_k = 1$$

Let $x_k$ stand for the number of people who choose the $k^\text{th}$ object when we sample the choices of $n$ people. Then the density function of ${x_k}$ is given by the multinomial pdf:

$$f(x_1,...x_K|-) = \begin{cases} \frac{n!}{x_1! ... x_K!} p_1^{x_1} ... p_K^{x_K} \quad \text{if} \quad \sum_kx_k=n \\ 0 \quad \text{otherwise}\end{cases}$$

You can then use maximum likelihood theory to estimate $\{p_1,p_2,...p_K\}$ and compute confidence intervals for these estimates.

Computing the confidence intervals would also enable you to compute margin of errors associated with your estimates for a given sample size. These margin of errors will help you calculate the necessary sample sizes to attain a margin of error of 5% with 95% confidence.

MLE, Margin of error and Sample Size Computations

It is not difficult to show that the MLE estimate for $p_k$ is given by:

$$\hat{p}_k = \frac{x_k}{n}$$

The above estimator is unbiased as:

$$E(\hat{p}_k) = \frac{E(x_k)}{n} = \frac{n p_k}{n}=p_k$$

The variance of the estimator is:

$$V(\hat{p}_k) = \frac{V(x_k)}{n^2} = \frac{n p_k (1-p_k)}{n^2}=\frac{p_k (1-p_k)}{n}$$

Assuming that $n$ is sufficiently high, we can use the central limit theorem to approximate the distribution of $\hat{p}_k$ as a normal with the mean at $p_k$ and with variance $\frac{p_k (1-p_k)}{n}$.

Thus, the margin of error is given for a 95% confidence interval is given by:

$$1.96 \sqrt{\frac{p_k (1-p_k)}{n}}$$

We do not know $p_k$ apriori. However, a conservative estimate for $p_k$ would be that it equals $K^{-1}$ (i.e., we assume that all choices are equally likely). The above argument is a bit ad-hoc but perhaps serves the OP's purpose.

Therefore, we have the requirement that:

$$1.96 \sqrt{\frac{K^{-1} (1-K^{-1})}{n}} = 0.05$$

If we let $K=3$ we get the required sample size as $n=341.475$.

PS: The past question on Asymptotic distribution of multinomial seems relevant in the above context and may suggest ways to lend rigor to the above ideas.

You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ (i.e., $K$ represents the number of colors, answers etc).

Let $p_k$ be the true proportion of people in the population who would choose the $k^\text{th}$ choice when presented with $K$ choices. You can re-interpret $p_k$ as the probability that a random person would choose the $k^\text{th}$ choice when presented with $K$ choices. Thus, by definition, we have:

$$\sum_{k=1}^K p_k = 1$$

Let $x_k$ stand for the number of people who choose the $k^\text{th}$ object when we sample the choices of $n$ people. Then the density function of ${x_k}$ is given by the multinomial pdf:

$$f(x_1,...x_K|-) = \begin{cases} \frac{n!}{x_1! ... x_K!} p_1^{x_1} ... p_K^{x_K} \quad \text{if} \quad \sum_kx_k=n \\ 0 \quad \text{otherwise}\end{cases}$$

You can then use maximum likelihood theory to estimate $\{p_1,p_2,...p_K\}$ and compute confidence intervals for these estimates.

Computing the confidence intervals would also enable you to compute margin of errors associated with your estimates for a given sample size. These margin of errors will help you calculate the necessary sample sizes to attain a margin of error of 5% with 95% confidence.

MLE, Margin of error and Sample Size Computations

It is not difficult to show that the MLE estimate for $p_k$ is given by:

$$\hat{p}_k = \frac{x_k}{n}$$

The above estimator is unbiased as:

$$E(\hat{p}_k) = \frac{E(x_k)}{n} = \frac{n p_k}{n}=p_k$$

The variance of the estimator is:

$$V(\hat{p}_k) = \frac{V(x_k)}{n^2} = \frac{n p_k (1-p_k)}{n^2}=\frac{p_k (1-p_k)}{n}$$

Assuming that $n$ is sufficiently high, we can use the central limit theorem to approximate the distribution of $\hat{p}_k$ as a normal with the mean at $p_k$ and with variance $\frac{p_k (1-p_k)}{n}$.

Thus, the margin of error is given for a 95% confidence interval is given by:

$$1.96 \sqrt{\frac{p_k (1-p_k)}{n}}$$

We do not know $p_k$ apriori. However, a conservative estimate for $p_k$ would be that it equals $K^{-1}$ (i.e., we assume that all choices are equally likely). The above argument is a bit ad-hoc but perhaps serves the OP's purpose.

Therefore, we have the requirement that:

$$1.96 \sqrt{\frac{K^{-1} (1-K^{-1})}{n}} = 0.05$$

If we let $K=3$ we get the required sample size as $n=341.475$.

PS: The past question on Asymptotic distribution of multinomial seems relevant in the above context and may suggest ways to lend rigor to the above ideas.

You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ (i.e., $K$ represents the number of colors, answers etc).

Let $p_k$ be the true proportion of people in the population who would choose the $k^\text{th}$ choice when presented with $K$ choices. You can re-interpret $p_k$ as the probability that a random person would choose the $k^\text{th}$ choice when presented with $K$ choices. Thus, by definition, we have:

$$\sum_{k=1}^K p_k = 1$$

Let $x_k$ stand for the number of people who choose the $k^\text{th}$ object when we sample the choices of $n$ people. Then the density function of ${x_k}$ is given by the multinomial pdf:

$$f(x_1,...x_K|-) = \begin{cases} \frac{n!}{x_1! ... x_K!} p_1^{x_1} ... p_K^{x_K} \quad \text{if} \quad \sum_kx_k=n \\ 0 \quad \text{otherwise}\end{cases}$$

You can then use maximum likelihood theory to estimate $\{p_1,p_2,...p_K\}$ and compute confidence intervals for these estimates.

Computing the confidence intervals would also enable you to compute margin of errors associated with your estimates for a given sample size. These margin of errors will help you calculate the necessary sample sizes to attain a margin of error of 5% with 95% confidence.

MLE, Margin of error and Sample Size Computations

It is not difficult to show that the MLE estimate for $p_k$ is given by:

$$\hat{p}_k = \frac{x_k}{n}$$

The above estimator is unbiased as:

$$E(\hat{p}_k) = \frac{E(x_k)}{n} = \frac{n p_k}{n}=p_k$$

The variance of the estimator is:

$$V(\hat{p}_k) = \frac{V(x_k)}{n^2} = \frac{n p_k (1-p_k}{n^2}=\frac{p_k (1-p_k)}{n}$$

Assuming that $n$ is sufficiently high, we can use the central limit theorem to approximate the distribution of $\hat{p}_k$ as a normal with the mean at $p_k$ and with variance $\frac{p_k (1-p_k)}{n}$.

Thus, the margin of error is given for a 95% confidence interval is given by:

$$1.96 \sqrt{\frac{p_k (1-p_k)}{n}}$$

We do not know $p_k$ apriori. However, a conservative estimate for $p_k$ would be that it equals $K^{-1}$ (i.e., we assume that all choices are equally likely). The above argument is a bit ad-hoc but perhaps serves the OP's purpose.

Therefore, we have the requirement that:

$$1.96 \sqrt{\frac{K^{-1} (1-K^{-1})}{n}} = 0.05$$

If we let $K=3$ we get the required sample size as $n=341.475$.

You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ (i.e., $K$ represents the number of colors, answers etc).

Let $p_k$ be the true proportion of people in the population who would choose the $k^\text{th}$ choice when presented with $K$ choices. You can re-interpret $p_k$ as the probability that a random person would choose the $k^\text{th}$ choice when presented with $K$ choices. Thus, by definition, we have:

$$\sum_{k=1}^K p_k = 1$$

Let $x_k$ stand for the number of people who choose the $k^\text{th}$ object when we sample the choices of $n$ people. Then the density function of ${x_k}$ is given by the multinomial pdf:

$$f(x_1,...x_K|-) = \begin{cases} \frac{n!}{x_1! ... x_K!} p_1^{x_1} ... p_K^{x_K} \quad \text{if} \quad \sum_kx_k=n \\ 0 \quad \text{otherwise}\end{cases}$$

You can then use maximum likelihood theory to estimate $\{p_1,p_2,...p_K\}$ and compute confidence intervals for these estimates.

Computing the confidence intervals would also enable you to compute margin of errors associated with your estimates for a given sample size. These margin of errors will help you calculate the necessary sample sizes to attain a margin of error of 5% with 95% confidence.

MLE, Margin of error and Sample Size Computations

It is not difficult to show that the MLE estimate for $p_k$ is given by:

$$\hat{p}_k = \frac{x_k}{n}$$

The above estimator is unbiased as:

$$E(\hat{p}_k) = \frac{E(x_k)}{n} = \frac{n p_k}{n}=p_k$$

The variance of the estimator is:

$$V(\hat{p}_k) = \frac{V(x_k)}{n^2} = \frac{n p_k (1-p_k)}{n^2}=\frac{p_k (1-p_k)}{n}$$

Assuming that $n$ is sufficiently high, we can use the central limit theorem to approximate the distribution of $\hat{p}_k$ as a normal with the mean at $p_k$ and with variance $\frac{p_k (1-p_k)}{n}$.

Thus, the margin of error is given for a 95% confidence interval is given by:

$$1.96 \sqrt{\frac{p_k (1-p_k)}{n}}$$

We do not know $p_k$ apriori. However, a conservative estimate for $p_k$ would be that it equals $K^{-1}$ (i.e., we assume that all choices are equally likely). The above argument is a bit ad-hoc but perhaps serves the OP's purpose.

Therefore, we have the requirement that:

$$1.96 \sqrt{\frac{K^{-1} (1-K^{-1})}{n}} = 0.05$$

If we let $K=3$ we get the required sample size as $n=341.475$.

PS: The past question on Asymptotic distribution of multinomial seems relevant in the above context and may suggest ways to lend rigor to the above ideas.

You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ (i.e., $K$ represents the number of colors, answers etc).

Let $p_k$ be the true proportion of people in the population who would choose the $k^\text{th}$ choice when presented with $K$ choices. You can re-interpret $p_k$ as the probability that a random person would choose the $k^\text{th}$ choice when presented with $K$ choices. Thus, by definition, we have:

$$\sum_{k=1}^K p_k = 1$$

Let $x_k$ stand for the number of people who choose the $k^\text{th}$ object when we sample the choices of $n$ people. Then the density function of ${x_k}$ is given by the multinomial pdf:

$$f(x_1,...x_K|-) = \begin{cases} \frac{n!}{x_1! ... x_K!} p_1^{x_1} ... p_K^{x_K} \quad \text{if} \quad \sum_kx_k=n \\ 0 \quad \text{otherwise}\end{cases}$$

You can then use maximum likelihood theory to estimate $\{p_1,p_2,...p_K\}$ and compute confidence intervals for these estimates.

Computing the confidence intervals would also enable you to compute margin of errors associated with your estimates for a given sample size. These margin of errors will help you calculate the necessary sample sizes to attain a margin of error of 5% with 95% confidence.

You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ (i.e., $K$ represents the number of colors, answers etc).

Let $p_k$ be the true proportion of people in the population who would choose the $k^\text{th}$ choice when presented with $K$ choices. You can re-interpret $p_k$ as the probability that a random person would choose the $k^\text{th}$ choice when presented with $K$ choices. Thus, by definition, we have:

$$\sum_{k=1}^K p_k = 1$$

Let $x_k$ stand for the number of people who choose the $k^\text{th}$ object when we sample the choices of $n$ people. Then the density function of ${x_k}$ is given by the multinomial pdf:

$$f(x_1,...x_K|-) = \begin{cases} \frac{n!}{x_1! ... x_K!} p_1^{x_1} ... p_K^{x_K} \quad \text{if} \quad \sum_kx_k=n \\ 0 \quad \text{otherwise}\end{cases}$$

You can then use maximum likelihood theory to estimate $\{p_1,p_2,...p_K\}$ and compute confidence intervals for these estimates.

Computing the confidence intervals would also enable you to compute margin of errors associated with your estimates for a given sample size. These margin of errors will help you calculate the necessary sample sizes to attain a margin of error of 5% with 95% confidence.

MLE, Margin of error and Sample Size Computations

It is not difficult to show that the MLE estimate for $p_k$ is given by:

$$\hat{p}_k = \frac{x_k}{n}$$

The above estimator is unbiased as:

$$E(\hat{p}_k) = \frac{E(x_k)}{n} = \frac{n p_k}{n}=p_k$$

The variance of the estimator is:

$$V(\hat{p}_k) = \frac{V(x_k)}{n^2} = \frac{n p_k (1-p_k}{n^2}=\frac{p_k (1-p_k)}{n}$$

Assuming that $n$ is sufficiently high, we can use the central limit theorem to approximate the distribution of $\hat{p}_k$ as a normal with the mean at $p_k$ and with variance $\frac{p_k (1-p_k)}{n}$.

Thus, the margin of error is given for a 95% confidence interval is given by:

$$1.96 \sqrt{\frac{p_k (1-p_k)}{n}}$$

We do not know $p_k$ apriori. However, a conservative estimate for $p_k$ would be that it equals $K^{-1}$ (i.e., we assume that all choices are equally likely). The above argument is a bit ad-hoc but perhaps serves the OP's purpose.

Therefore, we have the requirement that:

$$1.96 \sqrt{\frac{K^{-1} (1-K^{-1})}{n}} = 0.05$$

If we let $K=3$ we get the required sample size as $n=341.475$.