
statistics help 2
When the BC Government introduced a carbon tax in 2008, it did not want to lose votes as might easily happen with a new tax. They therefore guaranteed that revenue from the carbon tax would be channeled back into reducing income taxes. In 2011 Environics asked 1023 British Columbians whether they supported the tax and found 54% in favour.
What is the probability of getting this survey result (or higher) if in fact only 48% of British Columbians supported the tax.
Please can someone direct me towards the best way to approach this problem

Re: statistics help 2
I got P = 0.0001 after I found the z value of 0.9999. Is this the correct answer? It seems very small.

Re: statistics help 2
I also got that answer.
$\displaystyle H_0 : p = p_0$ vs $\displaystyle H_1: p \neq p_0 $
$\displaystyle Z = \frac{pp_0}{\sqrt{p_0(1p_0)/n}} = 3.84 $
We want $\displaystyle P ( 3.84 < Z < 3.84) $
And the p value on R 2*pnorm(abs(3.84)) = 0.000123.
So you STRONGLY reject the null hypothesis.