SOLUTION: Dear tutor please help me with stats questions, A marketing manager at a big retail shop claims that the variance in the weights of cereal boxes are less than 0.03 kilogram square

Question 1195940: Dear tutor please help me with stats questions, A marketing manager at a big retail shop claims that the variance in the weights of cereal boxes are less than 0.03 kilogram square. Thus, the weights (in kilograms) of a random sample of eight of these cereal boxes are listed here.
1.07; .98; .95; 1.05; .99; 1.09; 1.03; .96
a) Test this clam at a 1% level of significance.
b) Estimate the variance of the entire population of cereal box weights with 90% confidence.
Answer by ElectricPavlov(122) (Show Source): You can put this solution on YOUR website!

**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.

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