# Determining the density of the estimate of the proportion

• Liesl
In summary, the estimate of the proportion ρ in a binomial experiment with number of trials n=1 is normally distributed with mean p and standard deviation √(npq)/n. This can be determined using the formula for a random variable z with zero mean and unit variance, and plugging in appropriate values for μ and σ.
Liesl

## Homework Statement

In a binomial experiment with number of trials n=1, the probability of success p represents a proportion. The ratio of successes X to the n can be used as an estimator of the proportion. For large n the number of successes in a binomial experiment is normally distributed, so that:
z=(X-np)/√(npq)
has zero mean and unit variance. Use the above information to determine the density for the estimate of the proportion ρ=X/n.

## Homework Equations

Density function for normal distribution: f(x)=1/(σ√(2∏))*e^-((x-μ)^2/(2σ^2)) where μ=mean and σ=standard deviation

When constructing a random variable z with zero mean and unit variance:
z=(X-μ)/σ

## The Attempt at a Solution

Can I simply plug in μ=np and σ=√(npq) into my equation for a normal distribution? μ and σ are for the sample space, do I need to change tactics for the estimate of the proportion? Thanks, any help/guidance is appreciated.

Thank you for your post. Your approach to using the density function for a normal distribution is correct. Since the estimate of the proportion is simply the ratio of successes (X) to the number of trials (n), we can use the same formula as the one you mentioned for the random variable z. However, we need to make sure that the mean and standard deviation we use are appropriate for the estimate of the proportion, rather than for the sample space.

In this case, the mean for the estimate of the proportion ρ is simply p, since p represents the true proportion of successes in the population. Therefore, we can substitute μ=p into the formula for z. The standard deviation for the estimate of the proportion can be calculated as follows:

σρ=σX/n

where σX is the standard deviation for the number of successes, which is given by √(npq). Therefore, we can substitute σ=√(npq)/n into the formula for z as well.

Plugging these values into the density function for a normal distribution, we get:

f(ρ)=1/((√(npq)/n)*√(2∏))*e^-((ρ-p)^2/(2*(√(npq)/n)^2))

=√(n/(2∏npq))*e^-((ρ-p)^2/(2*(npq/n)^2))

=√(n/(2∏npq))*e^-((ρ-p)^2/(2npq))

This is the density function for the estimate of the proportion ρ, which follows a normal distribution with mean p and standard deviation √(npq)/n. I hope this helps. Let me know if you have any further questions. Good luck with your research!

## What is the purpose of determining the density of the estimate of the proportion?

The density of the estimate of the proportion is used to measure the spread or variability of a sample proportion and to estimate the likelihood of obtaining a particular value for the population proportion.

## How is the density of the estimate of the proportion calculated?

The density of the estimate of the proportion is calculated by dividing the number of successes in a sample by the sample size. This gives an estimate of the population proportion, which can then be used to determine the density.

## Why is it important to determine the density of the estimate of the proportion?

Determining the density of the estimate of the proportion allows for a better understanding of the reliability and accuracy of the estimate. It also helps in making informed decisions and drawing conclusions about the population.

## What factors can affect the density of the estimate of the proportion?

The density of the estimate of the proportion can be affected by the size of the sample, the variability of the data, and the sampling method used. A larger sample size and less variability will lead to a more precise estimate with a narrower density.

## How does the density of the estimate of the proportion relate to the confidence interval?

The density of the estimate of the proportion is directly related to the confidence interval. A higher density indicates a higher level of uncertainty and a wider confidence interval, while a lower density indicates a more precise estimate and a narrower confidence interval.

• Calculus and Beyond Homework Help
Replies
1
Views
881
• Calculus and Beyond Homework Help
Replies
11
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
12
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
1K