To determine whether the variance in the cereal box fillings exceeds the design specification, you need to perform a hypothesis test for variance using the Chi-Square distribution. Here’s how you can approach this:

Step 1: Set up the hypotheses.

• Null hypothesis ($H_0$): The population variance is equal to or less than the design specification.

  $H_0: \sigma^2 \leq 1$ ounces².

• Alternative hypothesis ($H_1$): The population variance is greater than the design specification.

  $H_1: \sigma^2 > 1$ ounces².

This is a one-tailed test (right-tailed).

Step 2: Identify the test statistic.

The test statistic for variance uses the Chi-Square ($\chi^2$) formula:

$\chi^2 = \frac{(n - 1) s^2}{\sigma^2_0}$

Where:

• $n$ is the sample size.
• $s^2$ is the sample variance (square of the sample standard deviation).
• $\sigma^2_0$ is the population variance under the null hypothesis.

Step 3: Calculate the test statistic.

You need the following:

• The sample size $n$.
• The sample standard deviation $s$.

Once you have these values, you can substitute them into the formula to calculate the test statistic.

Step 4: Find the critical value.

Use the Chi-Square distribution table (Table 11.1) to find the critical value for your chosen significance level $\alpha$ (often 0.05) and degrees of freedom $df = n - 1$.

Step 5: Conclusion.

• If the calculated Chi-Square statistic exceeds the critical value from the table, you reject the null hypothesis and conclude that the population variance exceeds the design specification.
• If the calculated statistic is less than the critical value, you fail to reject the null hypothesis, meaning there is insufficient evidence to suggest that the variance exceeds the specification.

Please provide the sample size and the sample standard deviation so we can compute the test statistic.