Answer:
a) Mean = 6.5, sample standard deviation = 3.50
b) Standard error = 0.7826
c) Point estimate = 6.5
d) Confidence interval: (5.1469 ,7.8531)
Explanation:
We are given the following data set for students to earn bachelor's degrees.
4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15
a) Formula:
where
are data points,
is the mean and n is the number of observations.
Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

b) Standard Error

c) Point estimate for the mean time required for all college is given by the sample mean.

d) 90% Confidence interval:
Putting the values, we get,
e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.