Question

Required information The sugar content in a one-cup serving of a certain breakfast cereal was measured for a sample of 150 servings. The average was 11.9 g and the standard deviation was 1.1 g. Find a 95% upper confidence bound for the mean sugar content. Round the answer to two decimal places. The 95% upper confidence bound is Required information The sugar content in a one-cup serving of a certain breakfast cereal was measured for a samplo of 150 servings. The average was 11.9 g and the standard deviation was 11 g. The claim is mode that the mean sugat content is greater than 11.707 g. With what level of confidence can this statement be made? Express the answer as a percent and round to two decimal ploces. The level of confidence is

Answer 1

To find a 95% upper confidence bound for the mean sugar content, calculate the Z-score, find the standard error, and use the formula: upper bound = sample mean + (Z-score * standard error). The result is approximately 12.05 g.

Step-by-step explanation:

To find a 95% upper confidence bound for the mean sugar content, we can use the formula:

Upper bound = Sample Mean + (Z-score * Standard Error)

First, we need to find the Z-score for a 95% confidence level. Since it is a one-tailed test and we want an upper bound, the Z-score is 1.645.

The standard error can be calculated by dividing the standard deviation by the square root of the sample size:

Standard Error = Standard Deviation / √(Sample Size)

Plugging in the given values, the standard error is:

Standard Error = 1.1 / √(150) ≈ 0.09

Now we can calculate the upper bound:

Upper bound = 11.9 + (1.645 * 0.09) ≈ 12.05 g

Therefore, the 95% upper confidence bound for the mean sugar content is approximately 12.05 g.