The international olympic committe states that the female participation in the 2004 summer olympic games was 42%, even with new sports such as weight lifting, hammer throw, and modern pentathalon being added to the games. Broadcasting and clothing companies want to change their advertising and marketing strategies if the female participation increases at the next games. An independent sports expert reported that 202 of 454 athletes in the random sample were women. is this strong evidence that the participation rate may increase?

Question

The international olympic committe states that the female participation in the 2004 summer olympic games was 42%, even with new sports such as weight lifting, hammer throw, and modern pentathalon being added to the games.  Broadcasting and clothing companies want to change their advertising and marketing strategies if the female participation increases at the next games. An independent sports expert reported that 202 of 454 athletes in the random sample were women. is this strong evidence that the participation rate may increase?
1) test an appropriate hypothesis and state your conclusion.
2) clearly explain what a Type-I error is in this context and give a possible consequence of making a type-I error.
3) clearly explain what a type II error is in this context and give a possible consequence of making a type-II error.
4) clealry explain, in context, what your P-value means

MathGuru · Answer

You can try a proportional one-sample z-test for this one since this problem is using proportions.

Here's a few hints to get you started:

Null hypothesis:
Ho: p = .42 -->meaning: population proportion is equal to .42
Alternative hypothesis:
Ha: p > .42 -->meaning: population proportion is greater than .42

Using a formula for a proportional one-sample z-test with your data included, we have:
z = .44 - .42 -->test value (202/454 is approximately .44) minus population value (.42) divided by
√[(.42)(.58)/454] --> .58 represents 1-.42 and 454 is sample size.

Finish the calculation. Remember if the null is not rejected, then there is no difference. If you need to find the p-value for the test statistic, check a z-table. The p-value is the actual level of the test statistic.

I'll give you some background on Type I and Type II errors and let you take it from there.

Type I errors result when you reject the null and it's true. Type II errors result when you accept the null and it's false. You can reduce Type I errors by setting the alpha at a lower level, for example, from .05 to .01. However, when you do that, you increase the probability of making a Type II error. You would have to determine if the interested parties would be more concerned about Type I or Type II errors.

I hope these hints will help.