Question

A random sample of 100 undergraduate students at a university found that 78 of them had used the university library’s website to find resources for a class. What is the margin of error for the true proportion of all undergraduates who had used the library’s website to find resources for a class?

Answer 1

margin of error suggests a confidence interval, but no width has been suggested

Answer 2

if we wanted to be 99% confident that the true population proportion was within a given interval, we would use z=2.576 and use the standard error as a multiplier

Answer 3

what would the equation I would use be???

Answer 4

Hint: use the central limit theorem, and use a maximum value for the standard deviation for a bernouli random variable.

Answer 5

Now, as amistre indicated, we don't know how "confident" you want the estimate to be, but you can leave it unspecified, e.g., the alpha percentile.

Answer 6

the standard error is just a modification of the standard deviation.

in this case: and i dont really understand why, the binomial standard deviation is sqrt(npq) with a mean of np (the mean i get)

\[z=\frac{x-np}{\sqrt{npq}}\]
\[z=\frac{\frac{x-np}{n}}{\frac{\sqrt{npq}}{n}}\]
\[z=\frac{\frac{x}{n}-p}{ \sqrt{\frac{npq}{n^2}}}\]
\[z=\frac{\frac{x}{n}-p}{ \sqrt{\frac{pq}{n}}}\]
\[z \sqrt{\frac{pq}{n}}=\frac{x}{n}-p\]
\[\underbrace{p}_{midpoint}\pm \underbrace{z \sqrt{\frac{pq}{n}}}_{margin~of~error}=\underbrace{\pm\frac{x}{n}}_{end~points}\]

Answer 7

something like that,