In: Statistics and Probability
The Business Education Department has six full-time faculty members. At their monthly department meeting, each full-time faculty member reported the number of hours they taught during the past month:
Faculty Member Teaching Hours
Bernhardt 84, Haas 92, England 84, Pecord 83, Hines 92, Tanner 83
A. Calculate the population mean for the six full-time faculty members:
B. If two full-time faculty members are selected randomly, how many different samples are possible?
C. Calculate the sample means:
D. Calculate the sampling distribution:
(A)
Population Mean () is given by:
So,
Answer is:
86.3333
(B)
Number of different samples = 6C2 = 15
So,
Answer is:
15
(C)
The sample means are got as follows:
Sample | Mean |
84,92 | 88 |
84,84 | 84 |
84,83 | 83.5 |
84,92 | 88 |
84,83 | 83.5 |
92,84 | 88 |
92,83 | 87.5 |
92,92 | 92 |
92,83 | 87.5 |
84,83 | 83.5 |
84,92 | 88 |
84,83 | 83.5 |
83,92 | 87.5 |
83,83 | 83 |
92,83 | 87.5 |
(D)
The Sampling Distribution is got as follows:
n = 15
Mean of Sampling Distribution of Sample means is given by
x | x- | (x-)2 |
88 | 1.6667 | 2.7778 |
84 | -2.3333 | 5.4444 |
83.5 | -2.8333 | 8.0278 |
88 | 1.6667 | 2.7778 |
83.5 | -2.8333 | 8.0278 |
88 | 1.6667 | 2.7778 |
87.5 | 1.1667 | 1.3611 |
92 | 5.6667 | 32.1111 |
87.5 | 1.1667 | 1.3611 |
83.5 | -2.8333 | 8.0278 |
88 | 1.6667 | 2.7778 |
83.5 | -2.8333 | 8.0278 |
87.5 | 1.1667 | 1.3611 |
83 | -3.3333 | 11.1111 |
87.5 | 1.16667 | 1.3611 |
Total = | 97.3333 |
Standard Deviation of Sampling Distribution of Sample means is
given by
So
Answer is:
The Sampling Distribution of Sample means is Normal Distribution with mean = 86.3333 and Standard Deviation = 2.5473