For this study, we should use z-test for a population proportion Correct The null and alternative hypotheses would be:

Question

10% of all students volunteer. Is the percentage of students who are volunteers different from students receiving financial aid? Of the 359 randomly selected students who receive financial aid, 39 of them volunteered. what can be concluded at the alpha = 0.10 level of significance?

using z-test for a population mean.

the null and alternative hypotheses would be:

Ho: p = 0.10 H1: p = 0.10

The test statistic z =???????(I don't know how to find this)

The p-value = ???????(or this)

The p-value is > than alpha

Based on this we should fail to reject the null hypothesis.

Answer 1

You are well on your way to the answer!

H0: p = .10 correct.

but H1 should be, H1: p ≠ p.10

1.) What quantities are being asked about: the sampling distribution of a proportion. A proportion is simply a special case of a mean, and you may use the statistical tests appropriate for the sampling distribution of a mean.

2.) You have a large N here, so you use a single sample Z test instead of a t test.

3.) The Z score is calculated as follows:

Z = (proportion observed in sample vs hypothesized population portion) / standard error of the proportion

4.) You need to go put a few pieces together to get the parts for your Z score:

1. calculate your sample proportion, 39/359 = .1086
2. calculate your standard error (same calculation as standard error of the mean)
3. The variance of a proportion is p*(1-p), your case, (.1086)(.8914)
4. The standard error is sqrt(variance/N)

5.) Look up your Z score in a z-table or run it through excels NORM.S.DIST() function.

1. You are doing a two-tailed test because you have been asked whether or not there is a "difference" between two quantities, not whether one is greater than the other, or whether one is less than the other.
2. You will need to find the p(Z <=) the negative value of your Z score + the p(Z >=) the positive value of your Z score. For instance, if you calculated a Z score of -1.58, you would want to find out the p(Z) <= 1.58 , and the p(Z) >= 1.58. The good news is you only have to do one of them and then multiply the probability by 2. This is your p-value.
3. If you are using excel, enter =NORM.S.DIST(your z score, 1). (The "1" tells excel you are working with the cumulative distribution). This will give you p(Z <= Z). Do 1 - p(Z <= Z) to get the p(Z >= z) for a positive z value. Again, 2 * the probability you get as an answer will be your p-value

If your p value < .10, reject H0.