Question 1193999
**1. Set up Hypotheses**

* **Null Hypothesis (H₀):** σ² ≥ 1.2² (The population variance of the technician's measurements is greater than or equal to 1.44) 
* **Alternative Hypothesis (H₁):** σ² < 1.2² (The population variance of the technician's measurements is less than 1.44)

**2. Determine the Test Statistic**

* We will use the Chi-Square test statistic for testing the population variance.
* **Test Statistic:** χ² = (n - 1) * s² / σ₀² 
    * where:
        * n = sample size (20 measurements)
        * s² = sample variance (1.1² = 1.21)
        * σ₀² = hypothesized population variance (1.2² = 1.44)

* **Calculate the Test Statistic:**
    * χ² = (20 - 1) * 1.21 / 1.44 = 19 * 1.21 / 1.44 = 15.96

**3. Determine the Critical Value**

* **Degrees of Freedom (df):** df = n - 1 = 20 - 1 = 19
* **Significance Level (α):** α = 0.05 
* **Critical Value:** 
    * Since this is a left-tailed test, we find the critical value from the Chi-Square distribution table with 19 degrees of freedom and an area of 0.05 in the left tail. 
    * Using a Chi-Square table or statistical software, the critical value for χ²(0.05, 19) is approximately 8.907.

**4. Decision**

* **Compare Test Statistic to Critical Value:**
    * Our calculated test statistic (15.96) is greater than the critical value (8.907).

* **Decision:** 
    * Since the test statistic does not fall in the rejection region (less than the critical value), we **fail to reject the null hypothesis**.

**5. Summary Statement**

* There is **not enough evidence** at the 0.05 significance level to conclude that the lab technician is meeting the target accuracy of a standard deviation less than 1.2. 

**Key Points:**

* We use the Chi-Square test for testing hypotheses about population variance.
* The critical value is determined based on the significance level and degrees of freedom.
* We compare the test statistic to the critical value to make a decision about the null hypothesis.