Recently, a university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates returned the survey. The average salary was $55,000. The population standard deviation is known to be $2,500. What is the 90% confidence interval for the mean salary of all graduates from the English Department?
A) 5,497, $55,039
B) 54,794, $55,206
C) 54,671, $55,329
D) 54,961, $5,539
E) 54,800, $55,200

Question

Quizplus · Accepted Answer

We will use the formula: 
Confidence interval = sample mean ± (z-score x (population standard deviation / square root of sample size))
Since the sample size is large (n = 400), we can assume normal distribution and use the z-score for a 90% confidence level, which is 1.645.

Confidence interval = 55,000 ± (1.645 x (2,500 / √400))
Confidence interval = 55,000 ± (1.645 x 125)
Confidence interval = 55,000 ± 205.625
Confidence interval = (55,000 - 205.625, 55,000 + 205.625)
Confidence interval = (54,794.375, 55,205.625)

Therefore, the answer is B, 54,794 to $55,206.

Recently, a University Surveyed Recent Graduates of the English Department