Question 1195940

**1. Calculate Sample Statistics**

* **Sample Mean (x̄):** 
   * Calculate the average of the given weights. 
   * x̄ = (1.07 + 0.98 + 0.95 + 1.05 + 0.99 + 1.09 + 1.03 + 0.96) / 8 
   * x̄ = 8.12 / 8 = 1.015 kg

* **Sample Variance (s²):**
   * Calculate the variance of the sample using the formula:
      * s² = Σ(x - x̄)² / (n - 1) 
      * where:
         * x: Individual weight
         * x̄: Sample mean
         * n: Sample size (8)
   * s² ≈ 0.0023 kg²

**2. State Hypotheses**

* **Null Hypothesis (H₀):** σ² = 0.03 (Variance of population weights is 0.03 kg²)
* **Alternative Hypothesis (H₁):** σ² < 0.03 (Variance of population weights is less than 0.03 kg²)

**3. Determine Test Statistic**

* **Chi-Square Test Statistic:**
   * χ² = (n - 1) * s² / σ₀² 
      * where:
         * n: Sample size (8)
         * s²: Sample variance (0.0023)
         * σ₀²: Hypothesized population variance (0.03)

   * χ² = (8 - 1) * 0.0023 / 0.03 
   * χ² = 0.5333

**4. Determine Critical Value**

* **Degrees of Freedom:** df = n - 1 = 8 - 1 = 7
* **Significance Level:** α = 0.01 (1%)
* **Chi-Square Distribution Table:** 
   * Find the critical value (χ²_critical) from the chi-square distribution table with 7 degrees of freedom and α = 0.01. 
   * χ²_critical ≈ 2.167 (for a one-tailed test)

**5. Decision Rule**

* **Reject H₀ if χ² < χ²_critical**

**6. Make a Decision**

* Since our calculated χ² (0.5333) is less than the critical value (2.167), we **reject the null hypothesis**.

**Conclusion:**

* There is sufficient evidence at the 1% level of significance to support the claim that the variance in the weights of cereal boxes is less than 0.03 kg².

**b) Estimate the Variance of the Population with 90% Confidence**

* **Confidence Level:** 90%
* **Degrees of Freedom:** df = n - 1 = 7
* **Find Chi-Square Values:**
    * Find the chi-square values (χ²_lower and χ²_upper) from the chi-square distribution table for 7 degrees of freedom and 5% and 95% significance levels (since it's a two-tailed interval).
      * χ²_lower ≈ 2.167 
      * χ²_upper ≈ 14.067

* **Calculate Confidence Interval:**
    * Lower Bound: (n - 1) * s² / χ²_upper = 7 * 0.0023 / 14.067 ≈ 0.00114
    * Upper Bound: (n - 1) * s² / χ²_lower = 7 * 0.0023 / 2.167 ≈ 0.0075

**Conclusion:**

* We are 90% confident that the true variance of the population of cereal box weights lies between 0.0011 kg² and 0.0075 kg².

**Note:**

* This analysis assumes that the weights of the cereal boxes are normally distributed. 
* The chi-square distribution is used to estimate the population variance.
**a) Hypothesis Testing**

* **Hypotheses:**
    * **Null Hypothesis (H0):** σ² ≥ 0.03 (Variance of cereal box weights is greater than or equal to 0.03 kg²)
    * **Alternative Hypothesis (H1):** σ² < 0.03 (Variance of cereal box weights is less than 0.03 kg²)

* **Test Statistic:** 
    * We will use the chi-square test statistic: 
        * χ² = (n - 1) * s² / σ₀² 
        * where:
            * n = sample size (8)
            * s² = sample variance
            * σ₀² = hypothesized population variance (0.03)

* **Calculate Sample Variance (s²)**

1. Calculate the sample mean (x̄):
   * x̄ = (1.07 + 0.98 + 0.95 + 1.05 + 0.99 + 1.09 + 1.03 + 0.96) / 8 = 1.015

2. Calculate the squared deviations from the mean:
   * (1.07 - 1.015)² = 0.002925
   * (0.98 - 1.015)² = 0.001225
   * ... and so on for all data points

3. Sum the squared deviations.

4. Divide the sum of squared deviations by (n - 1) = 7 to get the sample variance (s²). 
   * s² ≈ 0.001486

* **Calculate Test Statistic:**

   * χ² = (8 - 1) * 0.001486 / 0.03 
   * χ² ≈ 0.347

* **Determine Critical Value:**

   * Find the critical value of chi-square (χ²_critical) for a left-tailed test with α = 0.01 and degrees of freedom (df) = n - 1 = 7.
   * Use a chi-square distribution table or a statistical software. 
   * For α = 0.01 and df = 7, χ²_critical ≈ 2.167

* **Decision Rule:**

   * If the calculated χ² is less than the critical value (χ² < χ²_critical), we fail to reject the null hypothesis.
   * If the calculated χ² is greater than or equal to the critical value (χ² ≥ χ²_critical), we reject the null hypothesis.

* **Conclusion:**

   * Since 0.347 < 2.167, we fail to reject the null hypothesis. 
   * There is not enough evidence at the 1% significance level to support the claim that the variance in the weights of cereal boxes is less than 0.03 kg².

**b) Estimate the Variance with 90% Confidence**

* **Find Confidence Interval Bounds**

   * The confidence interval for the population variance is given by:

      * [(n - 1) * s² / χ²_upper, (n - 1) * s² / χ²_lower] 

      * where:
          * χ²_upper and χ²_lower are the upper and lower critical values of the chi-square distribution for (n - 1) degrees of freedom and the desired confidence level (90%).

* **Find Critical Values:**

   * For a 90% confidence level, α/2 = 0.05.
   * Use a chi-square distribution table or software to find:
      * χ²_upper (for α/2 = 0.05 and df = 7) 
      * χ²_lower (for 1 - α/2 = 0.95 and df = 7)

* **Calculate Confidence Interval:**
   * Substitute the values of n, s², χ²_upper, and χ²_lower into the formula to calculate the lower and upper bounds of the confidence interval for the population variance.

**Note:**

* The specific values of χ²_upper and χ²_lower will depend on the chi-square distribution table or software used. 
* This analysis assumes that the weights of the cereal boxes are normally distributed.

This approach provides a method for estimating the population variance with 90% confidence.