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Let's solve the problem step-by-step:

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### (a) **Level of significance and hypotheses**

- **Level of significance (\(\alpha\))**:  
  It is stated in the problem as **0.05**.

- **Null hypothesis (\(H_0\))**:  
  The null hypothesis assumes that the variance of the new shot is equal to or larger than the variance of the old shot:  
  \[
  H_0: \sigma^2 \geq 3^2
  \]

- **Alternative hypothesis (\(H_1\))**:  
  The alternative hypothesis states that the variance of the new shot is smaller than the variance of the old shot:  
  \[
  H_1: \sigma^2 < 3^2
  \]

---

### (b) **Chi-square test statistic and degrees of freedom**

- The formula for the chi-square test statistic is:  
  \[
  \chi^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2}
  \]  
  where:
  - \(n = 25\) (sample size),
  - \(s^2 = 2.1^2 = 4.41\) (sample variance),
  - \(\sigma_0^2 = 3^2 = 9\) (population variance under \(H_0\)).

  Substituting:
  \[
  \chi^2 = \frac{(25-1) \cdot 4.41}{9} = \frac{24 \cdot 4.41}{9} = 11.76
  \]

- **Degrees of freedom (\(df\))**:  
  \(df = n - 1 = 25 - 1 = 24\).

- **Assumptions about the population distribution**:  
  The chi-square test assumes a **normal population distribution**.

---

### (c) **P-value of the test statistic**

Using a chi-square table or calculator, find the **P-value** corresponding to \(\chi^2 = 11.76\) with \(df = 24\).  

- The P-value for this test is greater than **0.10**, as \(\chi^2 = 11.76\) is far to the left in the chi-square distribution for 24 degrees of freedom.

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### (d) **Decision regarding the null hypothesis**

The decision rule is:  
- Reject \(H_0\) if \(P\)-value \(\leq \alpha\).  
- Fail to reject \(H_0\) if \(P\)-value > \(\alpha\).

Here, \(P\)-value > 0.10, which is greater than the level of significance (\(\alpha = 0.05\)).  
**Conclusion**: We fail to reject the null hypothesis.

---

### (e) **Interpretation of the conclusion**

Since we fail to reject \(H_0\):  
**There is insufficient evidence to conclude that the new typhoid shot has a smaller variance of protection times.**

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### Summary of Answers

(a)  
- Level of significance: **0.05**  
- \(H_0: \sigma^2 \geq 9\)  
- \(H_1: \sigma^2 < 9\)  

(b)  
- \(\chi^2 = 11.76\)  
- Degrees of freedom: **24**  
- Assumption: **Normal population distribution**  

(c)  
- \(P\)-value > 0.10  

(d)  
- Fail to reject \(H_0\).  

(e)  
- There is insufficient evidence to conclude that the new typhoid shot has a smaller variance of protection times.  

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Would you like further explanation of any step?  

### Related Questions:
1. How is the chi-square test statistic calculated for two-sided tests?
2. What are the assumptions required for the chi-square test for variance?
3. Can the chi-square test be applied for small sample sizes? Why or why not?
4. How does changing the level of significance affect the conclusion?
5. What are alternative methods to test for differences in variances?  

**Tip**: Always ensure the population data approximately follows a normal distribution before applying the chi-square test for variance.

A new typhoid shot is being developed by a medical research team. The old typhoid shot was known to protect the population for a mean time of 36 months, with a standard deviation of 3 months. To test the time variability of the new shot, a random sample of 25 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 2.1 months. Using a 0.05 level of significance, test the claim that the new typhoid shot has a smaller variance of protection times.

Learn how to perform a chi-square test for variance with a step-by-step solution. This example evaluates the effectiveness of a new typhoid shot by testing its variance against a known standard using hypothesis testing at a 0.05 significance level.

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