Determining the density of the estimate of the proportion

[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.

 

[mighty2000]

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!