A college admissions officer takes a simple random sample of 103 entering freshmen and computes their mean mathematics SAT score to be 493. Assume the population standard deviation is 112. What is the lower bound of the 99% confidence interval?
Round your answer to the nearest integer.
99% = mean ± 2.575 SEm
SEm = SD/√n
I'll let you do the calculations.
To find the lower bound of the 99% confidence interval, we can use the formula:
Lower bound = sample mean - (Z-value * standard error)
Here, the sample mean is 493, and the population standard deviation is 112.
To calculate the Z-value for a 99% confidence interval, we need to find the critical value from the Z-table.
Since we want to find the lower bound, we need to find the Z-value associated with the probability of 0.01 in the Z-table, as 1 - 0.99 = 0.01 represents the lower tail area.
By looking up the Z-table, the Z-value associated with a 0.01 probability is approximately -2.33.
Now, we need to calculate the standard error:
Standard error = population standard deviation / square root of the sample size
Standard error = 112 / square root of 103
Standard error ≈ 11.02 (rounded to two decimal places)
Plugging these values into the formula:
Lower bound = 493 - (-2.33 * 11.02)
Lower bound ≈ 519.97
Rounded to the nearest integer, the lower bound of the 99% confidence interval is 520.