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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.

Respuesta :

somia2105
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)

ashimasood1990
95% confidence interval is ( 1340.06 , 1459.94)   
   ( solution is attached )
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