Posted by lola on .
From a random sample of 500 males interviewed, 125 indicated that they watch professional football on Monday night television. Does this evidence indicate that more than 20% of the male TV viewers watch professional football on Monday evenings? Use the 0.01 level of significance, and find the pvalue.

statistics 
MathGuru,
You can try a proportional onesample ztest for this one since this problem is using proportions.
Here's a few hints to get you started:
Null hypothesis:
Ho: p = .20 >meaning: population proportion is equal to .20
Alternative hypothesis:
Ha: p > .20 >meaning: population proportion is greater than .20
Using a formula for a proportional onesample ztest with your data included, we have:
z = .25  .20 >test value (125/500 is .25) minus population value (.20) divided by
√[(.25)(.75)/500] > .75 represents 1.25 and 500 is sample size.
Finish the calculation. Remember if the null is not rejected, then you cannot conclude p > .20. If you need to find the pvalue for the test statistic, check a ztable, then compare to .01 to determine whether or not to reject the null.
I hope this helps. 
Correction 
MathGuru,
This section should read:
Using a formula for a proportional onesample ztest with your data included, we have:
z = .25  .20 >test value (125/500 is .25) minus population value (.20) divided by
√[(.20)(.80)/500] > .80 represents 1.20 and 500 is sample size.
Sorry for any confusion.