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To verify the students' claim that those who take the course in Filipino obtain a higher average rating compared to those who take it in English, we can perform a hypothesis test using the given data. The test can be conducted at a 95% confidence level (which corresponds to a 5% significance level). Let's work through the steps outlined in the problem.

### Given Data:
- **English:**
  - Mean (\(\bar{x}_1\)) = 86
  - Standard Deviation (\(s_1\)) = 4
  - Sample Size (\(n_1\)) = 45
- **Filipino:**
  - Mean (\(\bar{x}_2\)) = 89
  - Standard Deviation (\(s_2\)) = 5
  - Sample Size (\(n_2\)) = 50

### 1) Hypotheses:
- **Null Hypothesis (\(H_0\))**: \(\mu_1 = \mu_2\) (The average ratings for English and Filipino courses are the same).
- **Alternative Hypothesis (\(H_1\))**: \(\mu_2 > \mu_1\) (The average rating for the Filipino course is higher than that for the English course).

### 2) Level of Significance:
- \( \alpha = 0.05 \) (corresponds to a 95% confidence level).

### 3) Test Statistic:
Since the sample sizes are large (both \(n_1\) and \(n_2\) are greater than 30), we can use a **two-sample z-test** for comparing means. The test statistic \(z\) is given by:

\[
z = \frac{\bar{x}_2 - \bar{x}_1}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]

Substituting the values:

\[
z = \frac{89 - 86}{\sqrt{\frac{4^2}{45} + \frac{5^2}{50}}}
\]

\[
z = \frac{3}{\sqrt{\frac{16}{45} + \frac{25}{50}}}
\]

\[
z = \frac{3}{\sqrt{0.3556 + 0.5}}
\]

\[
z = \frac{3}{\sqrt{0.8556}} \approx \frac{3}{0.9240} \approx 3.247
\]

### 4) Decision:
Now, compare the calculated \(z\)-value with the critical \(z\)-value for a one-tailed test at \( \alpha = 0.05 \).

- The critical \(z\)-value from the standard normal distribution table is approximately **1.645**.

Since \(z = 3.247 > 1.645\), we reject the null hypothesis.

### 5) Conclusion:
There is sufficient evidence at the 95% confidence level to support the claim that students who take the course in Filipino have a higher average rating than those who take it in English.

---

Would you like further details on any part of this solution, or do you have any questions? Here are some related questions to consider:

1. What assumptions must be met for the two-sample z-test to be valid?
2. How would the conclusion change if the sample sizes were smaller?
3. What would be the effect of using a 99% confidence level instead of 95%?
4. How would we conduct this test if the population variances were unknown and unequal?
5. Can this test be conducted using a t-test instead of a z-test? If so, under what conditions?

**Tip:** Understanding the conditions and assumptions of hypothesis tests is crucial, as violating them can lead to incorrect conclusions.

A college offers courses in logic in either Filipino or English. Students argue that those who take the course in Filipino obtain a higher average rating. The following are the results of the study conducted: (Assume Normal Distribution)
- English: Mean = 86, SD = 4, n = 45
- Filipino: Mean = 89, SD = 5, n = 50
Verify the students' claim if they are 95% confident about it.

Learn how to verify students' claims using hypothesis testing with a two-sample z-test. This problem involves comparing mean ratings in Filipino and English logic courses, with a 95% confidence level. Suitable for undergraduate statistics courses.

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