SOLUTION: A marketing manager at a big retail shop claims that the variance in the weights of cereal boxes are less than 0.003 kilogram 2 . Thus, the weights (in kilograms) of a random sa

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Question 1195912: A marketing manager at a big retail shop claims that the variance in the weights of cereal boxes are less
than 0.003 kilogram
2
. 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
4.1 Test this clam at a 1 % level of significance (8 marks)
4.2 Estimate the variance of the entire population of cereal box weights with 90% confidence.
(7 marks)
Answer by ElectricPavlov(122) About Me  (Show Source):
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**4.1 Test this claim at a 1% level of significance**
**1. Hypotheses**
* **Null Hypothesis (H₀):** σ² ≥ 0.003 (Variance of cereal box weights is greater than or equal to 0.003 kg²)
* **Alternative Hypothesis (H₁):** σ² < 0.003 (Variance of cereal box weights is less than 0.003 kg²)
**2. Calculate Sample Statistics**
* **Sample Mean (x̄):**
* x̄ = (1.07 + 0.98 + 0.95 + 1.05 + 0.99 + 1.09 + 1.03 + 0.96) / 8 = 1.015 kg
* **Sample Variance (s²):**
* 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
* Sum the squared deviations.
* Divide the sum of squared deviations by (n - 1) = 7 to get the sample variance (s²)
* s² ≈ 0.001486 kg²
**3. Calculate Test Statistic**
* **Chi-Square Test Statistic:**
* χ² = (n - 1) * s² / σ₀²
* where:
* n: sample size (8)
* s²: sample variance (0.001486 kg²)
* σ₀²: hypothesized population variance (0.003 kg²)
* χ² = (8 - 1) * 0.001486 / 0.003
* χ² ≈ 3.46
**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.
* For a one-tailed test with α = 0.01 and df = 7, χ²_critical ≈ 2.167
**5. Decision Rule**
* **Reject H₀ if χ² < χ²_critical**
**6. Make a Decision**
* Since our calculated χ² (3.46) is greater than the critical value (2.167), we **fail to reject the null hypothesis**.
**7. Conclusion**
* There is not enough evidence at the 1% level of significance to support the claim that the variance in the weights of cereal boxes is less than 0.003 kg².
**4.2 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 * 0.001486 / 14.067 ≈ 0.000738
* Upper Bound: (n - 1) * s² / χ²_lower = 7 * 0.001486 / 2.167 ≈ 0.004809
**Conclusion:**
* We are 90% confident that the true variance of the population of cereal box weights lies between 0.000738 kg² and 0.004809 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.