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 Unit 2 Questions for Discussion 7 Answer Key 1) The sample standard deviation will be larger because the denominator is n − 1 instead of n. (you will be dividing by a smaller number and the missing multiple will be larger.) In the calculator, the sample standard deviation will be larger than the population standard deviation. This is because the sample standard deviation is calculated by dividing by a smaller number (n-­‐1), instead of n. The result is a larger standard deviation. This is because if your data is just a sample, it would otherwise underestimate the population in general so it is compensating. (Of course, if the data set you entered were really your entire population, you would divide by n only and the standard deviation would be smaller. See σ x on the calculator.) 2) The sample standard deviation will use an s when giving standard deviation. If it is the standard deviation of the population, the standard deviation will be denoted as σ (the lowercase Greek letter sigma.) 3) a) The range is 15-­‐2=13. To find the standard 56
deviation: first find the mean, x =
= 8. . Find each 7
deviation from the mean x − x and square each (
)
(
)
2
deviation x − x . See table at the right. Sum the squared deviations. The sum is 108. Divide by n − 1 108
= 18. 7 −1
Take the square root. 18 = 4.24 b) The range is 41-­‐22=19. To find the standard 210
deviation, first find the mean x =
= 30. Find each 7
(
)
(
)
2
deviation from the mean x − x and square each deviation x − x . See table at the right. Sum the squared deviations. The sum is 236. Divide by n − 1 236
= 39.3 7 −1
Take the square root. 39.3 ≈ 6.27 c) The range is 13-­‐1=12. To find the standard deviation, first find the mean. 150
mean x =
= 7.143 Find each deviation from the 21
(
)
(
)
2
mean x − x and square each deviation x − x . These steps are in the table. The fourth column shows the frequency of each value multiplied by the squared deviation. After multiplying each of these, find the sum: 260,562. Divide by n − 1 . Remember that the total number of the values is 21. 260,562
= 13.0281 . Take the square root 13.0281 ≈ 3.61 21− 1
4) 84.2 + 79.0
a) Find the mean of the three groups. Eastern teams: = 81.6, Central teams: 2
76.4(5) + 78.0(6)
86.0(4) + 84.0(5)
= 77.3, Western teams = 84.9. The teams in the West had the 11
9
greatest winning average. (Note that we must use a weighting factor, frequency, in the mean calculations for the Central and Western teams since there are different number of teams in each of these leagues.) 15.5 + 14.1
b) Find the means of the standard deviations. Eastern teams: = 14.8, Central Teams: 2
10.9(5) + 9.7(6)
9.0(4) + 10.9(5)
= 10.25 , Western Teams = 10.06. The Eastern teams were the 11
9
least “consistent” because their standard deviation is the largest. Note again that we must use a weighting factor in the Central and Western teams. c) The average number of games won for all West Division teams is 84.9. 5) The histogram on the left has the greater variance. The mean is at the center of the middle class in each histogram. In the histogram on the left, six of the nine observations are in classes far away from the mean. Only two out of nine observations are in classes far away from the mean in the histogram on the right. Thus the variance is greater for the histogram on the left. 6) Since the variance is the squared dispersion around the mean, it can never be less than zero. It is impossible to have less dispersion than no dispersion.