Question

Students in an introductory statistics class were asked to report the age of their mothers when they were born. Summary statistics include Sample size: 28 students Sample mean: 29.643 years Sample standard deviation: 4.564 years a. Calculate the standard error of this sample mean. b. Determine and interpret a 90% confidence interval for the mother’s mean age (at student’s birth) in the population of all students at this university. c. How would a 99% confidence interval compare to the 90% interval in terms of its midpoint and half-width? d. Would you expect 90% of the ages in the sample to be within the 90% confidence interval? Explain why or why not. e. Even if the distribution of mothers’ ages were somewhat skewed, would this confidence interval procedure still be valid with these data? Explain why or why not.

Answer 1

A) Standard Error of mean is 4.564/sqrt(28) = 0.8625

B)
Lower limit of confidence interval
mean - t(0.1, 27) * SE = 28.17413708 years
Upper limit of confidence interval
mean + t(0.1, 27) * SE = 31.11186292 years

we are 90% confident that the mother's age (at student's birth) lies in the interval from 28.17413708 to 31.11186292 years.

c)

z= 2.58

99 CI = {Mean +/- S.D. * Z/ sqrt(n)} = 29.643 +/- 2.58*4.564/ sqrt(28) = {27.418 , 31.868}

compare all and answer