Final answer:
The 99% confidence interval for the mean with n=12, x=28.0, and s=5.7 is (22.89, 33.11), after determining the t-value and calculating the margin of error, which corresponds to option E.
Therefore, option E is the correct answer.
Step-by-step explanation:
To find the 99% confidence interval for the mean when the sample mean (x) is 28.0, the sample standard deviation (s) is 5.7, and the sample size (n) is 12, you need to use the t-distribution because the sample size is less than 30. First, find the t-value that corresponds to the 99% confidence level with 11 degrees of freedom (which is n-1 for the sample size of 12). This value can be found in t-distribution tables or using statistical software. Once you have the t-value, you calculate the margin of error (ME) by multiplying the t-value by the standard deviation divided by the square root of the sample size (s/√n). Finally, add and subtract the margin of error from the sample mean to find the confidence interval.
Let's assume the appropriate t-value is approximately 3.106 based on a t-distribution table. Calculate the margin of error as follows:
ME = t * (s/√n)
ME = 3.106 * (5.7/√12)
ME = 3.106 * (5.7/3.4641)
ME = 3.106 * 1.6454
ME = 5.1143
Now, calculate the confidence interval:
Lower Limit = x - ME = 28.0 - 5.1143 = 22.8857
Upper Limit = x + ME = 28.0 + 5.1143 = 33.1143
After rounding to two decimal places, the confidence interval is (22.89, 33.11). Therefore, option E is the correct answer.