Question

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.

Answer 1

95% confidence interval is ( 1340.06 , 1459.94)
( solution is attached )

Answer 2

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)