# type 1 error and type 2 error

• Nov 18th 2010, 02:54 AM
lpd
type 1 error and type 2 error
A bowl contains seven marbles of which http://www.sosmath.com/CBB/latexrend...5dc8912759.gif are red while the others are blue. In order
to test the null hypothesis Ho : http://www.sosmath.com/CBB/latexrend...5dc8912759.gif = 2 against HI : http://www.sosmath.com/CBB/latexrend...5dc8912759.gif = 4, two of the marbles are
randomly drawn without replacement and the null hypothesis is rejected if and
only if both are red.

(a) What is the probability of committing a Type I error?
(b) What is the probability of committing a Type II error?

a) pr(2 red|theta=2) 2/7x1/6 = 1/21

b) pr(o red, 1 red|theta=4) = 3/7*2/6+4/7*3/6+3/7*4/6=5/7

can some1 explain this to me? i'm a bit lost. i know type 1 error is reject Ho when Ho is actually true.

for for a, i get 2 red marbles. so 2/7 and multiple it 1/6 chance? but why 1/6 chance?

and for b) type2 eror is failing to reject Ho when we actually should. just cant get my head around it some explaination would be great http://www.sosmath.com/CBB/images/smiles/icon_smile.gif

This is what i got so far
Type I error = "false positive" = rejecting H_0 when H_0 is true
Type II error = "false negative" = accepting H_0 when H_0 is false = accepting H_0 when H_1 is true

So, in symbols:
http://www.sosmath.com/CBB/latexrend...07be36ff57.gif
and
http://www.sosmath.com/CBB/latexrend...2aca9153ac.gif
Now you go back and plug in what is meant by reject/accept H_0, and the hypotheses H_0 and H_1 in this problem, to get
http://www.sosmath.com/CBB/latexrend...185b54fcb5.gif
and
http://www.sosmath.com/CBB/latexrend...a9c644c74f.gif
and then it is just a matter of calculating them, as in the solution.

then, are my numbers correct? just a bit lost in how to interpret the numbers within the question to get the answer.
• Nov 21st 2010, 09:53 PM
matheagle
alpha is (2/7)(1/6)=1/21
you can also do this by the hypergeo formula, but here we have to pick 2 red from a bowl with 2 reds and 5 blues.
You reach in and select a red with prob 2/7 and you DO NOT replace it.
Then there is 1 red left and still 5 whites, which has a 1/6 chance of being a red marble.
• Nov 21st 2010, 09:56 PM
matheagle
as for beta, you don't want 2 reds, which leads to rejection, so I would use the complement

$1-P(RR)=1-(4/7)(3/6)=5/7$ since we have 4 reds in our bowl now.