Question

1. (20 pts.) A hybrid course is one that contains both online and classroom instruction. In a study performed at Middle Georgia State College, a software package was used as the main source of instruction in a hybrid college algebra course. The software tracked the number of hours it took for each student to meet the objectives of the course. In a sample of 43 students, the mean number of hours was 80.3, with a standard deviation of 51.8.
   1. Construct a 99% confidence interval for the mean number of hours it takes for a student to meet the course objectives.

b. If a sample of 40 students had been studied, would you expect the interval to be wider or narrower than the interval constructed in part (a)? Explain

Answer 1

(a) Here sample size = n= 43

Mean number of hours = = 80.3 hours

sample standard deviation = s= 51.8 hours

standard error = s/sqrt(n) = 51.8/sqrt(43) = 7.8994 hours

we have to find 99% confidence interval

degree of freedom = dF = n-1 = 43 -1 = 42

Critical value = TINV(0.01, 42) = 2.6981

Margin of error = 2.6981 * 7.8994 = 21.3132 hours

99% confidence interval = 80.3 + 21.31 = (58.99 hours, 101.61 hours)

(b) If a sample of 40 to be studied, the 99% confidence interval is more wider than the interval constructed in part (a). It is because standard error will increase due to square root of sample mean in its formula in denominator.