Thirty random students took a calc test. If the sample mean was 90 & the standard deviation was 13.9, construct a 99% confidence interval for the mean score of all students.

Is the answer 83.03 to 96.97??

This looks within range. There may be differences due to rounding when using a confidence interval formula.

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

Confidence Interval = Sample Mean +/- (Critical Value * Standard Error)

Step 1: Find the critical value
The critical value depends on the desired level of confidence and the sample size. For a 99% confidence level and a sample size of 30, we can use a t-distribution and find the critical value from a t-table or calculator. In this case, the critical value is approximately 2.756.

Step 2: Calculate the standard error
The standard error measures the average deviation of sample means from the population mean. To calculate the standard error, divide the standard deviation by the square root of the sample size.

Standard Error = Standard Deviation / √(Sample Size)
Standard Error = 13.9 / √30
Standard Error ≈ 2.536

Step 3: Plug in the values into the formula
Using the formula, we can now construct the confidence interval:

Confidence Interval = Sample Mean +/- (Critical Value * Standard Error)
Confidence Interval = 90 +/- (2.756 * 2.536)

Calculating the confidence interval:
Confidence Interval = 90 +/- 7.008

Step 4: Calculate the confidence interval boundaries
To get the lower boundary, subtract the result from the sample mean. To get the upper boundary, add the result to the sample mean.

Lower Boundary = 90 - 7.008 ≈ 82.992
Upper Boundary = 90 + 7.008 ≈ 97.008

So, the 99% confidence interval for the mean score of all students is approximately 82.992 to 97.008.

Therefore, the answer you provided, 83.03 to 96.97, is very close to the correct answer based on rounding.