Thirty randomly selected students took the calculus final. If the sample mean was 89 and the standard deviation was 6.2, construct a 99% confidence interval of the mean score of all students. Assume that the population has a normal distribution

99% = mean ± Z*SEm

SEm = SD/√n

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.005) and its Z score. Insert Z score with other data in the first equation.

To construct a confidence interval for the mean score of all students, we can use the formula:

Confidence interval = sample mean ± (critical value) * (standard deviation / √n)

Here's how to calculate the confidence interval step by step:

Step 1: Determine the critical value.
Since we want a 99% confidence interval, we need to find the critical value that corresponds to a 99% confidence level. The critical value can be obtained from a standard normal distribution table or using a statistical calculator. For a 99% confidence level, the critical value is approximately 2.575.

Step 2: Calculate the standard error.
The standard error is calculated as the standard deviation divided by the square root of the sample size. In this case, the standard deviation is 6.2, and the sample size is 30. Therefore, the standard error is 6.2 / √30 ≈ 1.129.

Step 3: Calculate the confidence interval.
Using the formula from above, we can plug in the values:
Confidence interval = 89 ± (2.575) * (1.129)

Calculating the upper bound:
Upper bound = 89 + (2.575) * (1.129) = 89 + 2.907 = 91.907

Calculating the lower bound:
Lower bound = 89 - (2.575) * (1.129) = 89 - 2.907 = 86.093

Therefore, the 99% confidence interval for the mean score of all students is (86.093, 91.907). This means that there is a 99% probability that the true mean score of all students lies within this interval.