# Math Help - Type 1/ Type 2 Error Question

1. ## Type 1/ Type 2 Error Question

I have problems trying to construct a proper framework to deal with this one. Please help if you can!

A random sample of 500 registered voters in a rural city is asked if they favor abolition of road tax. If more than 400 voters respond positively, we will conclude that at least 60% of the voters favor the abolition of road tax.
a) Find the probability of Type 1 error if exactly 60% of the voters favor the abolition of road tax.
b) What is the type II error probability if 75% of the voters favor the abolition of road tax?

Thanks!

2. Originally Posted by chopet
I have problems trying to construct a proper framework to deal with this one. Please help if you can!

A random sample of 500 registered voters in a rural city is asked if they favor abolition of road tax. If more than 400 voters respond positively, we will conclude that at least 60% of the voters favor the abolition of road tax.
a) Find the probability of Type 1 error if exactly 60% of the voters favor the abolition of road tax.
b) What is the type II error probability if 75% of the voters favor the abolition of road tax?

Thanks!
H0: Less than 60% of the voters favor the abolition of road tax.

H1: More than 60% of the voters favor the abolition of road tax.

Let $\hat{p}$ represent the proportion of the sample that favor the abolition of road tax.

Then (for large n) $\hat{p}$ ~ Normal $\left( \mu = p, ~ \sigma = \sqrt{\frac{p(1-p)}{n}}\right)$

where p is the proportion of the population that favor the abolition of road tax and n is the sample size.

Probability of type I error = Pr(Reject H0 | H0 true) = $\Pr(\hat{p} > 0.8 | p = 0.6)$

$= \Pr(\hat{p} > 0.8 | \hat{p}$ ~ Normal $\left( \mu = 0.6, ~ \sigma = \sqrt{\frac{0.6(1-0.6)}{500}} = 0.022\right)$.

Probability of type II error = Pr(Don't reject H0 | H0 false) = $\Pr(\hat{p} < 0.8 | p = 0.75)$

$= \Pr( \hat{p} < 0.8 | \hat{p}$ ~ Normal $\left( \mu = 0.75, ~ \sigma = \sqrt{\frac{0.75(1-0.75)}{500}} = 0.019\right)$.

Calculations are left for you.

3. Thanks for the fast response.
Just to clarify:

For the Type 1 error:
All I have to do is calculate the z-value of 0.8 with a mean of 0.6:
${{0.8-0.6} \over {0.022} } = 9.09$

Won't that be way off the normal curve. Hence probability of Type 1 error is 0?

4. Originally Posted by chopet
Thanks for the fast response.
Just to clarify:

For the Type 1 error:
All I have to do is calculate the z-value of 0.8 with a mean of 0.6:
${{0.8-0.6} \over {0.022} } = 9.09$

Won't that be way off the normal curve. Hence probability of Type 1 error is 0?
Yes.

5. ## this values is true???

let x1,x2......xn be a random samole from N(mu,9) it is decided that the probability of rejecting :
mu=1 when mu=1.8 is (0.5)

and the probability of accepting mu<or equal 1 when mu=1.2 is (0.1)

if the test use :
mean(x bar)>K

find n and K:

solution :

i find this two unknowns and i get :

n= 304

and K=.799