Question 821573
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Hi,
A sample of 25 randomly English majors has a mean test score of 81.5 with a 
standard deviation of 10.2. 
ME = {{{1.96(10.2/sqrt(25))}}} = 3.9984
95% confidence interval for the population mean, &#956;.
  81.5 - 4< {{{mu}}} < 81.5 + 4