You wanted to test whether the cereal boxes made by your plant conform to the requirement that they contain 12 oz cereal. You wish to test at the 99 percent significance level, and you sample 100 boxes, finding the mean x as 11.85, standard deviation s is 0.5. Do your cereal boxes meet the standard?

Z = (score-mean)/SEm

SEm = SD/√n

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Is it significant?

To determine whether the cereal boxes meet the standard of containing 12 oz of cereal, we can conduct a hypothesis test using the given sample mean and standard deviation.

Let's define the null hypothesis (H₀) and alternative hypothesis (H₁) for this test:

H₀: The mean cereal content of the boxes is equal to 12 oz.
H₁: The mean cereal content of the boxes is not equal to 12 oz.

We will conduct a two-tailed test at the 99 percent significance level, which means we will reject the null hypothesis if the test statistic falls in either tail of the sampling distribution.

To perform this hypothesis test, we need the critical value associated with the 99 percent level of significance and the sample size.

Step 1: Determine the critical value (or rejection region).
Since we have a two-tailed test, we need to divide the significance level (α) by 2 to find the critical values for each tail.
The remaining significance level (α/2) for each tail is (1 - 0.99)/2 = 0.005.

Using a z-table or a statistical software, we can find the critical value for α/2 = 0.005, which is approximately ±2.58.

Step 2: Calculate the test statistic (z-score).
The test statistic (z) can be calculated using the formula:
z = (x - μ) / (s / sqrt(n))
where x is the sample mean, μ is the population mean (the standard of 12 oz in this case), s is the sample standard deviation, and n is the sample size.

Plugging in the values:
z = (11.85 - 12) / (0.5 / sqrt(100)) = -3 / 0.05 = -60

Step 3: Evaluate the test statistic.
Compare the absolute value of the test statistic (|-60| = 60) with the critical value (±2.58).
Since the test statistic falls far beyond the critical value, we can reject the null hypothesis.

Step 4: Conclusion.
Based on the results of the hypothesis test, we can conclude that the cereal boxes produced by your plant do not meet the standard of containing 12 oz of cereal.

Please note that the calculation above assumes that the cereal box weights are normally distributed and that the sample was selected randomly. It's also worth considering factors like the margin of error and practical significance when interpreting the results.