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 .25 ounces. Calculate a confidence interval to test the hypotheses that cereal weight is different from 13 ounces at α = 0.10 and interpret. Read more

Steps to Solve

1. Identify the sample data We have a sample size ($n$) of 31 boxes, a sample mean ($\bar{x}$) of 12.58 ounces, and a sample standard deviation ($s$) of 0.25 ounces.

2. Determine the level of confidence We need to calculate a 90% confidence interval which corresponds to a significance level ($\alpha$) of 0.10. For a two-tailed test, we therefore have $\alpha/2 = 0.05$ for each tail.

3. Find the critical t-value Using a t-table and degrees of freedom ($df = n - 1 = 30$), we find the critical value for a 90% confidence level. The critical t-value ($t_{0.05, 30}$) approximately equals 1.697.

4. Calculate the margin of error The margin of error (E) can be calculated with the formula: $$ E = t \times \frac{s}{\sqrt{n}} $$ Substituting the values: $$ E = 1.697 \times \frac{0.25}{\sqrt{31}} \approx 0.067 $$

5. Calculate the confidence interval The confidence interval can be calculated using: $$ \text{CI} = \bar{x} \pm E $$ Substituting the values: $$ \text{CI} = 12.58 \pm 0.067 $$ This results in the interval: $$ (12.513, 12.647) $$

6. Interpret the confidence interval Since the calculated confidence interval (12.513, 12.647) does not include the hypothesized mean of 13 ounces, we reject the null hypothesis. This implies that there is sufficient evidence at the 0.10 level to conclude that the average weight of the cereal boxes is significantly different from 13 ounces.