Step-by-step explanation:
1. **Null and Alternative Hypothesis:**
- Null Hypothesis (\(H_0\)): The percentage of college students using X remains at 60%.
- Alternative Hypothesis (\(H_1\)): The percentage of college students using X has changed from 60%.
2. **Test Statistic and P-value:**
To conduct a hypothesis test for proportions, we can use the z-test for proportions.
- Test Statistic:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]
Where:
- \(\hat{p}\) is the sample proportion (80/100 = 0.80)
- \(p_0\) is the hypothesized population proportion (0.60)
- \(n\) is the sample size (100)
- P-value:
We'll use the z-table or software to find the p-value corresponding to the calculated test statistic.
3. **Decision and Conclusion:**
- Decision Rule:
- If the p-value is less than the significance level (\(\alpha = 0.05\)), reject the null hypothesis.
- If the p-value is greater than or equal to the significance level, fail to reject the null hypothesis.
- Conclusion:
- If we reject the null hypothesis, we conclude that there is evidence to suggest that the percentage of college students using X has changed.
- If we fail to reject the null hypothesis, we conclude that there is not enough evidence to suggest a change in the percentage of college students using X.
To complete the analysis, we need to calculate the test statistic and the p-value using the given data. Let me do the calculations.
Calculating the test statistic:
\[
z = \frac{0.80 - 0.60}{\sqrt{\frac{0.60(1 - 0.60)}{100}}} = \frac{0.20}{\sqrt{\frac{0.60 \times 0.40}{100}}}
\]
\[
z = \frac{0.20}{\sqrt{\frac{0.24}{100}}} = \frac{0.20}{\sqrt{0.0024}} \approx \frac{0.20}{0.049}
\]
\[
z \approx 4.08
\]
Now, we need to find the p-value associated with \(z \approx 4.08\). This value represents the probability of observing a test statistic as extreme as \(z \approx 4.08\) under the null hypothesis.
Using a z-table or statistical software, we find that the p-value is extremely small, close to 0. This indicates strong evidence against the null hypothesis.
**Decision and Conclusion:**
Since the p-value (\(p \approx 0\)) is less than the significance level (\(\alpha = 0.05\)), we reject the null hypothesis. Therefore, we conclude that there is evidence to suggest that the percentage of college students using X has changed.
