Use the critical value approach to test the following, at the 5% significance level.
𝑥̅ = 30 𝑠 = 8 𝑛 = 15
Ho: μ = 20 Ha: μ  20
Test statistic: t  4.841
a) Reject the null hypothesis. At the 5% significance level, the data provide sufficient evidence that the
population mean does not equal 20

To solve this hypothesis testing problem using the **critical value approach**, let us work through the steps systematically:

---

### **1. Hypotheses**
- Null Hypothesis (\(H_0\)): \(\mu = 20\)
- Alternative Hypothesis (\(H_a\)): \(\mu \neq 20\) (two-tailed test)

---

### **2. Significance Level**
- \(\alpha = 0.05\) (5% significance level)

---

### **3. Test Statistic**
The test statistic provided is:
\[
t \approx 4.841
\]

---

### **4. Degrees of Freedom**
The degrees of freedom for a \(t\)-test is calculated as:
\[
df = n - 1 = 15 - 1 = 14
\]

---

### **5. Critical Value**
For a two-tailed test at \(\alpha = 0.05\) and \(df = 14\), we find the critical values from a \(t\)-distribution table (or calculator):

- \(t_{critical} = \pm 2.145\)

This means that:
- Reject \(H_0\) if \(t < -2.145\) or \(t > 2.145\)

---

### **6. Decision Rule**
Compare the test statistic \(t = 4.841\) with the critical values:
- Since \(4.841 > 2.145\), the test statistic falls in the **rejection region**.

---

### **7. Conclusion**
Reject \(H_0\). At the 5% significance level, the data provide sufficient evidence to conclude that the population mean \(\mu\) is not equal to 20.

---

Let me know if you'd like further clarifications or if you have any other questions!  

### **Follow-Up Questions**
1. What is the formula for the test statistic \(t\)?
2. How do we interpret the significance level \(\alpha\)?
3. What changes in the analysis if this were a one-tailed test?
4. How do degrees of freedom affect the critical value in a \(t\)-test?
5. How can the \(p\)-value approach be used for the same test?

### **Tip:**  
Always check if the sample size is large enough to assume normality when performing \(t\)-tests. For smaller samples, ensure the population is approximately normal.

Use the critical value approach to test the following, at the 5% significance level. 𝑥̅ = 30 𝑠 = 8 𝑛 = 15
Ho: μ = 20 Ha: μ ≠ 20
Test statistic: t ≈ 4.841
(a) Reject the null hypothesis. At the 5% significance level, the data provide sufficient evidence that the population mean does not equal 20.

Learn how to perform hypothesis testing using the critical value approach at a 5% significance level. This problem involves calculating the test statistic, finding the critical values, and making a conclusion about whether the population mean equals 20. Suitable for Grades 11-12 students studying hypothesis testing and t-tests.

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