Question

18% of all college students volunteer their time. is the proportion of college students who are volunteers larger for students receiving financial aid? of the 365 randomly selected students who receive financial aid, 84 of them volunteered their time. what can be concluded at the α = 0.05 level of significance?
a. for this study, we should use z - test for a population proportion
b. the null and alternative hypotheses would be:
$h_0$: p = 0.18 (please enter a decimal)
$h_1$: p > 0.18 (please enter a decimal)
c. the test statistic z = (please show your answer to 3 decimal places.)
d. the p - value = (please show your answer to 4 decimal places.)
e. the p - value is? α
f. based on this, we should select an answer the null hypothesis.
g. thus, the final conclusion is that...
the data suggest the population proportion is not significantly higher than 18% at α = 0.05, so there is sufficient evidence to conclude that the percentage of financial aid recipients who volunteer is equal to 18%.
the data suggest the populaton proportion is significantly higher than 18% at α = 0.05, so there is sufficient evidence to conclude that the percentage of financial aid recipients who volunteer is higher than 18%.
the data suggest the population proportion is not significantly higher than 18% at α = 0.05, so there is insufficient evidence to conclude that the percentage of financial aid recipients who volunteer is higher than 18%.

Explanation:

Step1: Calculate the sample proportion

The sample proportion \(\hat{p}=\frac{x}{n}\), where \(x = 84\) and \(n=365\). So \(\hat{p}=\frac{84}{365}\approx0.23\)

Step2: Calculate the test - statistic

The formula for the \(z\) - test statistic for a proportion is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)

Given \(p = 0.18\), \(\hat{p}=0.23\), \(n = 365\)

First, calculate \(\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.18\times(1 - 0.18)}{365}}=\sqrt{\frac{0.18\times0.82}{365}}=\sqrt{\frac{0.1476}{365}}\approx\sqrt{0.0004044}\approx0.0201\)

Then \(z=\frac{0.23 - 0.18}{0.0201}=\frac{0.05}{0.0201}\approx2.49\)

Step3: Calculate the p - value

Since this is a right - tailed test (\(H_1:p>0.18\)), the p - value is \(P(Z>z)\)

Using the standard normal distribution table or a calculator, \(P(Z > 2.49)=1 - P(Z\leqslant2.49)\)

From the standard normal table, \(P(Z\leqslant2.49)=0.9936\)

So the p - value \(=1 - 0.9936=0.0064\)

Step4: Compare p - value and \(\alpha\)

Given \(\alpha=0.05\), and the p - value \(=0.0064\)

Since \(0.0064<0.05\) (i.e., p - value \(<\alpha\))

Step5: Make a decision

We reject the null hypothesis \(H_0\)

Answer:

c. \(z = 2.49\)

d. \(p - value=0.0064\)

e. \(p - value<\alpha\)

f. reject

g. The data suggest the populaton proportion is significantly higher than \(18\%\) at \(\alpha = 0.05\), so there is sufficient evidence to conclude that the percentage of financial aid recipients who volunteer is higher than \(18\%\).