A cereal manufacturer is concerned that the boxes of cereal not be underfilled. Each box of cereal is supposed to contain 13 ounces of cereal. A random sample of 31 boxes is tested... A cereal manufacturer is concerned that the boxes of cereal not be underfilled. Each box of cereal is supposed to contain 13 ounces of cereal. A random sample of 31 boxes is tested. The average weight is 12.58 ounces and the standard deviation is .25 ounces. Calculate a confidence interval to test the hypothesis that cereal weight is different from 13 ounces at α = 0.10 and interpret. Read more

Steps to Solve

1. Gather information and parameters
   Identify the given values:

• Sample size ($n$) = 31
• Sample mean ($\bar{x}$) = 12.58 ounces
• Standard deviation ($s$) = 0.25 ounces
• Significance level ($\alpha$) = 0.10

2. Calculate the critical value
   Since we are testing for a two-tailed hypothesis, we need the critical values for a 90% confidence level, which corresponds to $\alpha/2 = 0.05$.
   Using a t-distribution table for $n - 1 = 30$ degrees of freedom:

• Critical value ($t_{0.05, 30}$) ≈ 1.697

3. Calculate the standard error (SE)
   The standard error of the mean can be calculated using:
   $$ SE = \frac{s}{\sqrt{n}} $$
   Substituting the values:
   $$ SE = \frac{0.25}{\sqrt{31}} \approx 0.045 $$

4. Calculate the margin of error (ME)
   The margin of error is found by multiplying the critical value by the standard error:
   $$ ME = t_{0.05, 30} \times SE $$
   Substituting the values:
   $$ ME = 1.697 \times 0.045 \approx 0.076 $$

5. Determine the confidence interval
   Now we can calculate the confidence interval:
   $$ CI = \bar{x} \pm ME $$
   Calculating both bounds:
   $$ CI_{lower} = 12.58 - 0.076 \approx 12.504 $$
   $$ CI_{upper} = 12.58 + 0.076 \approx 12.656 $$

6. Interpret the results
   The 90% confidence interval for the average cereal weight is approximately (12.504, 12.656). Since this interval does not include the value 13 ounces, we conclude that the average weight of cereal boxes is significantly different from the expected 13 ounces at $\alpha = 0.10$.