unfair die

kclark36 · #1 $Old$ June 30th, 2009, 09:37 AM

Here is a problem that I have been working on. I am mixed up on the different testing formulas. I included what i did so far after each question. Thanks for looking

It is suspected that a particular die used in a game of chance was not fair. It was thought that the 1-side was heacy and the 6-side was light. Data was collected concerning the number of observed 1-s when the die was rolled 5000 times. Let p1 equal the probability of rolling a one with that die. We will test the hypothesis H0 p1=1/6 against the alternative hypothesis H1 p1<1/6 (i don't understand why this is < either)

a. give the test statistic and critical region that has an alpha=0.01 significant level

test stat $Z=\frac{Y/5000-1/6}{\sqrt{(1/6*5/6)/5000}}$
crit. reg. $Z \le -z_{0.01}$

b. If y=775 is the observed number of ones in the 5000 rolls, calculate the value of the test statistic and state conclusion about the die

$Z=\frac{775/5000-1/6}{.0052704628}=-2.213594348$

-2.21<-2.36

since -2.21 is not less than -2.36, H0 fails to be rejected

c. find the p-value for this test

$P(Z \le z)=P(Z \le -2.21)=1-0.9864=0.0136=$ ? This part confuses me because of the negative test statistic. I don't know how to calculate the p-value and which way the equality should be facing.

Thanks

malaygoel · #2 $Old$ June 30th, 2009, 09:50 AM

Quote:

Originally Posted by kclark36 $View Post$

Here is a problem that I have been working on. I am mixed up on the different testing formulas. I included what i did so far after each question. Thanks for looking

It is suspected that a particular die used in a game of chance was not fair. It was thought that the 1-side was heacy and the 6-side was light. Data was collected concerning the number of observed 1-s when the die was rolled 5000 times. Let p1 equal the probability of rolling a one with that die. We will test the hypothesis H0 p1=1/6 against the alternative hypothesis H1 p1<1/6 (i don't understand why this is < either)

I think that since 1-side was heacy, it makes more probable to be 1 on the lower side.

kclark36 · #3 $Old$ June 30th, 2009, 08:59 PM

I am wondering if someone can tell me if this is how this problem should be worked out. I'm hoping that I'm finally getting the hang of this. The p-value really confuses me.

matheagle · #4 $Old$ June 30th, 2009, 10:21 PM

since your p-value, .0136, is greater than alpha, you fail to reject the null hypothesis.
DRAW a st normal curve and look at both your test stat and where the rejection region begins.
If the p-value is less than alpha, then your test statistic is in your rejection region.
Hence you reject the null hypothesis is this case.
Likewise if the p-value is large (in comparison to alpha) you fail to reject the null hypothesis.

kclark36 · #5 $Old$ July 1st, 2009, 09:50 AM

Is that the correct way to do the P-value? I only found a p-value example for alternative hypothesis H1 where p>p0. I'm also not sure about the negative and how that changes things. Thanks

matheagle · #6 $Old$ July 1st, 2009, 10:01 AM

Quote:

Originally Posted by kclark36 $View Post$

Is that the correct way to do the P-value? I only found a p-value example for alternative hypothesis H1 where p>p0. I'm also not sure about the negative and how that changes things. Thanks

If the alternative hypothesis has a less than sign,
then you accept the alternative hypothesisis for small values (negative in this case).
That makes the p-value equal to P(Z<number).

kclark36 · #7 $Old$ July 1st, 2009, 10:11 AM

So rather than $P(Z \le -2.21)$ it should be $P(Z<-2.21)=1-0.9864=0.136$ ?

The example I have for alternate hypothesis of $H_{1}: p_{1}>p_{0}$ showed a p value of $P(Z \ge z)$ , so that is why I assumed to just reverse the equality but the negative confused me. I can't find much material on finding the p-value.

$\Phi 2.21=0.09864$

matheagle · #8 $Old$ July 1st, 2009, 10:15 AM

They are the same.
This is a continuous distribution.

$P(X<a)=P(X\le a)=\int_{-\infty}^a f(x)dx$

Since $P(X=a)=0$

$P(X<a)+P(X=a)=P(X\le a)$