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How to find the variance with this information?

In a certain Uni, a random sample of 600 student showed that 486 had found "graduate" employment 6 months after graduating. Determine a 90% confidence interval for the proportion of students in graduate employment.

So 90% tells me that there's 5% on either side for each interval, the sample size is large so we use the normal distrib table, which the z value is found to be 2.575(linear interpolate between 2 values)

The mean is then easily calculated simply by 486/600

now the formula for the intervals is the mean + or - Z(standrad error)

we have then so far 486/600 + or - 2.575(standard error)

standard error is simply standard deviation/root n

how do i find the standard deviation?


forget it, ignore me
(edited 6 years ago)
Fairly sure you need to say your X∼B(600,0.81)X \sim B(600,0.81) then note that this distribution is "like" the normal distribution X∼N(μ,σ2)X \sim N(\mu, \sigma^2) where μ\mu and σ2\sigma^2 are the mean and variance of the binomial distribution. All due to DeMoivre-Laplace CLT.

Anyway, if you got it then you got it - bumping this thread off unanswered.
(edited 6 years ago)
Original post by RDKGames
Fairly sure you need to say your X∼B(600,0.81)X \sim B(600,0.81) then note that this distribution is "like" the normal distribution X∼N(μ,σ2)X \sim N(\mu, \sigma^2) where μ\mu and σ2\sigma^2 are the mean and variance of the binomial distribution. All due to DeMoivre-Laplace CLT.

Anyway, if you got it then you got it - bumping this thread off unanswered.


ye so i know it as the "standard error or proportion" which is the p(1-p)/n all rooted thanks anyway :biggrin:

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