~Let's say that 220 out of 430 randomly selected male students participate in at least one college sponsored athletic activity at Marist College, and 150 out of 320 randomly selected females do as well. Does this data suggest that a higher percentage of males are involved in athletic sponsored activities at Marist? Test this claim at .01 level of significance.

~Solution: Using a TI-83/84 calculator.
We let P₁ denote the population proportion of males who are involved and P₂ the population proportion of females involved. We are interested in the following hypothesis test:

H_o:  P₁=P₂ (the percentage of males is not higher)
H₁:  P₁>P₂ (the percentage of males is higher)

~Procedure:
1) Press STAT, go to Tests, down to menu 6 (2-PropZTest)
2) Enter x₁: 220,  n₁: 430,  x₂: 150,  n₂: 320
3) This is a right-tailed test, so choose P₁>P₂
4) Calculate, Enter

~Results in the display:
1) P₁>P₂ (alternate hypothesis)
2) z=1.1617 (test statistic)
3) P-value: .1227 (determines our decision)
4) .5116 (sample proportion for males)
    .4688 (sample proportion for females)
5) Pooled sample proportion (370/750)
6) Sample sizes

~Conclusion.: Since the P-value of .1227 is not less than the significance level of .01, we do not reject H_o. Stating the conclusion properly, we say:

"There is insufficient evidence to conclude that more males participate
  in sponsored athletic activities at Marist compared to females."

~Note: Confidence Intervals estimate the difference P₁-P₂ between two population proportions. Therefore, the difference of our sample proportions is at the center of the interval.

~Example: Using a TI-83/84, go to menu B under STAT, Tests
(2-PropZInt) and enter the necessary information.

~If we did this with the above problem for a 95% level of confidence, we would get the following confidence interval: (-.0294, .11514)

~Note: The margin of error, E, is found the same way we found it for one population (subtract the center value from .11514).