Question 1177451: At the Fredericton High School it is estimated that at most 25% of the students ride bikes to school. Does this seem to be a valid estimate if, in a random sample of 90 high school students, 20 are found to ride bikes to school? Use = 0.01. Find the p-value for this test.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! We're going to conduct a hypothesis test for a proportion.
Let $p$ be the true proportion of students who ride bikes to school.
1. **Hypotheses:**
* Null hypothesis: $p \le 0.25$ (The estimate is valid.)
* Alternative hypothesis: $p > 0.25$ (The estimate is not valid; the true proportion is higher.)
2. **Test statistic:**
We use the z-statistic for proportions:
```
z = \frac{\hat{p} - p_0}{\sqrt{p_0 (1 - p_0) / n}}
```
where $\hat{p} = \frac{20}{90} = \frac{2}{9}$ is the sample proportion, $p_0 = 0.25$ is the hypothesized proportion, and $n = 90$ is the sample size. This gives us
```
z = \frac{\frac{2}{9} - 0.25}{\sqrt{0.25 (1 - 0.25) / 90}} \approx -0.23.
```
3. **P-value:**
The p-value is the probability of observing a sample proportion as extreme as $\hat{p} = \frac{2}{9}$, assuming the null hypothesis is true. Since this is a right-tailed test, the p-value is
```
P(Z > -0.23) = 1 - P(Z \le -0.23) \approx 1 - 0.4090 = 0.5910.
```
4. **Conclusion:**
Since the p-value (0.5910) is greater than $\alpha = 0.01$, we fail to reject the null hypothesis. There is not enough evidence to conclude that the estimate of at most 25% of students riding bikes to school is not valid.
|
|
|
