Question

The sugar content in a one-cup serving of a certain breakfast cereal was measured for a sample of 145 servings. The average was 11.9 g and the standard deviation was 1.1 g.

Find a 95% confidence interval for the mean sugar content.

Answer 1

To find a 95% confidence interval for the mean sugar content in a one-cup serving of the breakfast cereal, use the formula: CI = Xbar ± Z*(σ/sqrt(n)). The 95% confidence interval is (11.7 g, 12.1 g).

Step-by-step explanation:

To find a 95% confidence interval for the mean sugar content in a one-cup serving of the breakfast cereal, we can use the formula:

CI = Xbar ± Z*(σ/sqrt(n))

Where:

• CI is the confidence interval
• Xbar is the sample mean (11.9 g)
• Z is the critical value (from the Z-table, for a 95% confidence level it is approximately 1.96)
• σ is the population standard deviation (1.1 g)
• n is the sample size (145 servings)

Plugging in the values:

CI = 11.9 ± 1.96*(1.1/sqrt(145)) = 11.9 ± 0.200

Therefore, the 95% confidence interval for the mean sugar content is (11.7 g, 12.1 g).