A cereal manufacturer is concerned that the boxes of cereal not be underfilled or overfilled. Each box of cereal is supposed to contain 13 ounces of cereal. A random sample of 31 b... A cereal manufacturer is concerned that the boxes of cereal not be underfilled or overfilled. 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 0.25 ounces. Calculate a confidence interval to test the hypothesis that the cereal weight is different from 13 ounces at α = 0.10 and interpret. Read more

Steps to Solve

1. Determine the sample statistics

We have a sample of ( n = 31 ) boxes with a sample mean ( \bar{x} = 12.58 ) ounces and a sample standard deviation ( s = 0.25 ) ounces.

2. Calculate the standard error

The standard error (SE) of the mean is calculated using the formula:

$$ SE = \frac{s}{\sqrt{n}} $$

Substituting the values:

$$ SE = \frac{0.25}{\sqrt{31}} \approx 0.045 $$

3. Find the critical value

For a confidence level of ( 1 - \alpha = 0.90 ) (since ( \alpha = 0.10 )), we use a t-distribution because the sample size is less than 30. The degrees of freedom (df) is ( n - 1 = 31 - 1 = 30 ).

Using a t-table or calculator, find the critical t-value for ( df = 30 ) at ( \alpha/2 = 0.05 ):

$$ t_{0.05, 30} \approx 1.697 $$

4. Calculate the confidence interval

The confidence interval is given by:

$$ \bar{x} \pm t \times SE $$

Calculating this:

$$ 12.58 \pm 1.697 \times 0.045 $$

Calculating the margin of error:

$$ ME \approx 0.076 $$

So the confidence interval is:

$$ (12.58 - 0.076, 12.58 + 0.076) \approx (12.504, 12.656) $$

5. Interpret the results

The confidence interval ( (12.504, 12.656) ) means we are 90% confident that the true average weight of the cereal boxes is between 12.504 ounces and 12.656 ounces. Since this interval does not include 13 ounces, we can conclude that there is significant evidence to suggest that the cereal weight is different from 13 ounces.