A random sample of 64 sat scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240. determine the "t" value for a 95% confidence interval for the mean sat score. round your answer to three decimal places.

To find t-critical value,

with n = 64 and a 95% confidence interval t-value is 1.998

Sample mean = 1400

Standard deviation = 240

Standard error of mean = s/√n

Standard error of mean = 240/√64

SE = 240/8

Standard error of the mean = 30

Confidence interval 1400-30 (1.998) and 1400 + 30 (1.998)

95% confidence interval (1340.06, 1459.94)