# Need help with p-value question

• Jul 22nd 2009, 08:02 AM
thejollywag
Need help with p-value question
The question:

"According to an exit poll in the 2006 Virginia senatorial election, 50.1% of the sample size 2011 reported voting for James Webb. Is this enough evidence to predict who won? Test that the population proportion who voted for Webb was 0.50 against the alternative that it differed from 0.50. Answer by

b) Stating hypotheses and checking assumptions for a large-sample test.
c) Reporting the P-value and interpreting it. (The test statistic equals 0.09.)"

My questions

Am I correct in stating that the null hypothesis is H0: p=.5 and HA: p does not equal .5 (i.e., the little not equal symbol)? If so, do I want to use a two-tailed test?

Second, using the z=(p-hat - p=.5)/se_null (which is sqrt of .5(.5)/2011), I did find that the z-score rounds to .09, which, according to my z-table, gives a p-value of .5359. First, are the z-table figures analogous to p-values, and second, and am I supposed to subtract .5359 from 1 to get the tail value, and then multiply that by 2? I've gone over the text repeatedly but can't seem to figure it out.

Third, if anybody knows the TI-83, I'm getting a different answer for the z-score by using 1-PropZTest (p0=.5, x=1005-1008 (I've tried them all, and none of them bring up .09) and n=2011). 1-PropZTest is option 5 under Stat/Tests.

If anybody can give me any guidance at all, I would appreciate it greatly.

Thanks!
• Jul 22nd 2009, 12:32 PM
testing a proportion
Quote:

Originally Posted by thejollywag
The question:

"According to an exit poll in the 2006 Virginia senatorial election, 50.1% of the sample size 2011 reported voting for James Webb. Is this enough evidence to predict who won? Test that the population proportion who voted for Webb was 0.50 against the alternative that it differed from 0.50. Answer by

b) Stating hypotheses and checking assumptions for a large-sample test.
c) Reporting the P-value and interpreting it. (The test statistic equals 0.09.)"

My questions

Am I correct in stating that the null hypothesis is H0: p=.5 and HA: p does not equal .5 (i.e., the little not equal symbol)? If so, do I want to use a two-tailed test?

Second, using the z=(p-hat - p=.5)/se_null (which is sqrt of .5(.5)/2011), I did find that the z-score rounds to .09, which, according to my z-table, gives a p-value of .5359. First, are the z-table figures analogous to p-values, and second, and am I supposed to subtract .5359 from 1 to get the tail value, and then multiply that by 2? I've gone over the text repeatedly but can't seem to figure it out.

Third, if anybody knows the TI-83, I'm getting a different answer for the z-score by using 1-PropZTest (p0=.5, x=1005-1008 (I've tried them all, and none of them bring up .09) and n=2011). 1-PropZTest is option 5 under Stat/Tests.

If anybody can give me any guidance at all, I would appreciate it greatly.

Thanks!

Hi there

You are correct in saying that you need a two tailed test. Now you want to test that the true proportion p0=0.5. Let X be the number of voters who voted for the candidate then X~N(np0,np0q0) where q0=1-p0 will approximate the binomial distribution that describes X. The test statistic is

Z = X-(np0) / ( sqrt(np0q0) )

but since the normal distribution is continous and the binomial is discrete you need to apply a continuity correction which involves adding or subtracting 0.5 from X. thus the test statistic is

Z = (X-0.5)-(np0) / ( sqrt(np0q0) )
=(1008-0.5)-(2011*0.5) / ( sqrt(2011*0.5*0.5) )
=0.008919=0.009

Where 1008 is obtained by finding 50.1% of 2011 and rounding up.

Did the question mean the test statistic is 0.009 and not 0.09?

Since your test is 2-tailed and say you want 95% confidence then look up the 97.5th percentile of the standard normal dis which is 1.96 and reject h0:po=0.5 if the mod(Z)>1.96. Since our test statistic is 0.009 then we cannot reject h0.

the p-value for z=0.009 is approximately 1.

The conditions for the normal approximation to work are np0 and nq0 are both >5 which they are.

I hope this helps and that I have not mis-understood your question.
• Jul 22nd 2009, 06:41 PM
thejollywag