In a simple random sample of 1000 people taken from Town A, 13% are senior citizens. In an independent simple random sample of 600 people taken from Town B, 17% are senior citizens. Is the percent of senior citizens different in the two towns? Or is this just chance variation?

Choose the closest estimated standard error of the difference between the percents of senior citizens in the two samples, Under the null hypothesis.

1 or 1.2 or 1.4 or 1.6 or 1.8 or 2?

Question

In a simple random sample of 1000 people taken from Town A, 13% are senior citizens. In an independent simple random sample of 600 people taken from Town B, 17% are senior citizens. Is the percent of senior citizens different in the two towns? Or is this just chance variation?

Choose the closest  estimated standard error of the difference between the percents of senior citizens in the two samples, Under the null hypothesis.

1 or 1.2 or 1.4 or 1.6 or 1.8  or 2?

amistre64 · Answer

$$z=\frac{mean1 - mean2}{\sqrt{\frac{s^2}{n1}+\frac{s^2}{n2}}}$$
this sounds familiar ...

amistre64 · Answer

the under part might be the standard error if im remembering it right

amistre64 · Answer

can you recall what that might be from? http://www.wolframalpha.com/input/?i=sqrt%28%28.13*.87%2F1000%29%2B%28.17*.83%2F600%29%29

anonymous · Answer

is not like this?

anonymous · Answer

so is 1.8

anonymous · Answer

agree?

amistre64 · Answer

standard error of the means ... which relates to proportions of course

anonymous · Answer

so do you agree?

amistre64 · Answer

id have to review some stuff first :)

anonymous · Answer

Its correct do not waste your time i checked it ;) Thank you

amistre64 · Answer

standard error of the mean:$$\frac{\sigma}{\sqrt{n}}$$
using proportions that would amount to::$$\sqrt{\frac{pq}{n}}$$
the difference between means tends to have an addition under there;  and prolly by luck i got 1.8... as well

anonymous · Answer

the easiest way is always the best ;)