Question

An education researcher is interested in studying student performance on the SAT. The researcher takes a random sample of 200 students and calculates a sample mean of 500 and a sample standard deviation of 70.

19.   Calculate the standard error of the mean.

Answer: ____________________

21.   Calculate the 99% confidence interval.

Answer: ____________________

Answer 1

n = sample size = 200

sample mean = = 500

sample standard deviation = s = 70

19) Standard error( SE) of the mean is as

Answer : 4.949747

21) degrees of freedom = 200 - 1 = 199

t critical value for c = 0.99 so that   = 1 - c = 0.01 using excel is as

t_c = "=TINV(0.01,199)" = 2.60

Margin of error = t_c *SE = 2.60*4.949747 = 12.869

Lower limit = - E = 500 - 12.869 = 487.131

Upper limit = + E = 500 + 12.869 = 512.869

Answer: ( 487.131 ,  512.869 )