Question

Filling boxes with consistent amounts of its cereals is critical to General Mills's success. The filling variance for boxes of Count Chocula cereal is designed to be 0.02 ounces2 or less. A sample of 41 boxes of Count Chocula shows a sample standard deviation of 0.14 ounces. Use

α = 0.05 to determine whether the variance in the cereal box fillings is exceeding the design specification.

State the null and alternative hypotheses.
H0: σ2 > 0.02
Ha: σ2 ≤ 0.02
H0: σ2 ≥ 0.02
Ha: σ2 < 0.02
H0: σ2 < 0.02
Ha: σ2 ≥ 0.02
H0: σ2 = 0.02
Ha: σ2 ≠ 0.02
H0: σ2 ≤ 0.02
Ha: σ2 > 0.02

Find the value of the test statistic.

Answer 1

The correct hypotheses for the test of variance for General Mills's cereal box fillings are H0: σ2 ≤ 0.02 and Ha: σ2 > 0.02. The test statistic can be calculated using the provided data and compared against the chi-square distribution to determine if the variance exceeds the design specification.

Step-by-step explanation:

The question pertains to conducting a hypothesis test for the variance of a normally distributed population. The problem involves General Mills and their concern about whether the variance in the cereal box fillings is exceeding the design specification. The appropriate hypotheses for a test using variance would be the one where the null hypothesis (H0) represents no effect or no difference, which in this case means the variance is not greater than the design specification. Therefore, the null hypothesis (H0) should reflect the status quo, and the alternative hypothesis (Ha) should reflect the suspicion or what we are trying to detect.

The correct set of hypotheses would be:

• H0: σ2 ≤ 0.02 (The variance is less than or equal to the design specification of 0.02 ounces².)
• Ha: σ2 > 0.02 (The variance is greater than the design specification of 0.02 ounces².)

To calculate the test statistic, we would use the chi-square distribution with the formula:

χ² = (n-1)s² / σ2

Where n is the sample size, s is the sample standard deviation, and σ2 is the population variance under the null hypothesis. Substituting the given values we get:

χ² = (41-1)0.14² / 0.02

After computing the test statistic, we would compare it with the critical value from the chi-square distribution table at the significance level of α = 0.05. If the test statistic exceeds the critical value, we reject the null hypothesis indicating that the variance is indeed greater than the design specification.