Suppose that in a sample of 28 Moraine Valley students, the mean amount of time

Question

Suppose that in a sample of 28 Moraine Valley students, the mean amount of time they say they spend studying for classes each week is 4.1 hours with a standard deviation of 1.3 hours. a) Calculate a 95% confidence interval for the mean amount of time all Moraine Valley students spend studying for classes each week. b) Interpret your interval. c) What type of interval did you use for part (a) (z-interval, t-interval, 1-prop-z-interval) Why? d) What is the margin of error associated with your estimate?

Explanation / Answer

a) here std error of mean =std deviation/(n)1/2 =1.3/(28)1/2 =0.2457

for (n-1=27) degree of freedom and 95% confidence level; t=2.0518

therefore 95% confidence interval =sample mean -/+ t*std error =3.5959 ; 7.6959

b) above interval gives 95% confidence to contain true mean population amount of time Morain Valley student spent studying for classes.

c)as we do not know population std deviation therefore we should use t- interval

d) margin of error =t*std error =0.5041

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