$51 \%$ of students entering four-year colleges receive a degree within six years. Is this percent larger than for students who play intramural sports? 148 of the 264 students who played intramural sports received a degree within six years. What can be concluded at the level of significance of $\alpha=0.01$ ?
a. For this study, we should use z-test for a population proportion
b. The null and alternative hypotheses would be:
c. The test statistic $2+$ (please show your answer to 3 decimal places.)
of
d. The $p$-value $=$ (Please show your answer to 4 decimal places.)
e. The p-value is
f. Based on this, we should fail to reject the null hypothesis.
g. Thus, the final conclusion is that ...
The data suggest the populaton proportion is significantly larger than $51 \%$ at $\alpha=0.01$, so there is sufficient evidence to conclude that the population proportion of students who played intramural sports who received a degree within six years is larger than $51 \%$
The data suggest the population proportion is not significantly larger than $51 \%$ at $\alpha=0.01$, so there is not sufficient evidence to conclude that the population proportion of students who played intramural sports who received a degree within six years is larger than 51\%.
The data suggest the population proportion is not significantly larger than $51 \%$ at $\alpha=0.01$, so there is sufficient evidence to conclude that the population proportion of students who played intramural sports who received a degree within six years is equal to $51 \%$.
h. Interpret the p-value in the context of the study.
Answer
Interpret the p-value in the context of the study: The p-value of 0.1000 means that if the true population proportion of students who played intramural sports and received a degree within six years is 51%, there is a 10% chance of observing a sample proportion as extreme as the one we observed (or more extreme) just by random chance.
Step 1 ：For this study, we should use z-test for a population proportion.
Step 2 ：The null and alternative hypotheses would be: \(H_0: p = 0.51\) and \(H_1: p \neq 0.51\).
Step 3 ：The test statistic is calculated as (sample proportion - population proportion) / standard error. The standard error is calculated as \(\sqrt{(population proportion * (1 - population proportion)) / sample size}\). The test statistic \(z = 1.645\).
Step 4 ：The p-value is the probability of observing a test statistic as extreme as the one we calculated, assuming the null hypothesis is true. The p-value is \(0.1000\).
Step 5 ：The p-value is larger than the significance level of 0.01.
Step 6 ：Based on this, we should fail to reject the null hypothesis.
Step 7 ：Thus, the final conclusion is that the data suggest the population proportion is not significantly larger than \(51 \%\) at \(\alpha=0.01\), so there is not sufficient evidence to conclude that the population proportion of students who played intramural sports who received a degree within six years is larger than 51\%.
Step 8 ：Interpret the p-value in the context of the study: The p-value of 0.1000 means that if the true population proportion of students who played intramural sports and received a degree within six years is 51%, there is a 10% chance of observing a sample proportion as extreme as the one we observed (or more extreme) just by random chance.
