Problem Solution 2
College of Doctoral Studies
PSY 845: Module 2 Problem Set Solutions
Problem 1:
Find the mean, median, and mode for the scores in the following frequency distribution table:
(See pp. 99-100 for directions in your text book)
X f
8 1
7 4
6 2
5 2
4 2
3 1
Solution:
1. Transform information so that all scores appear in the order of their values. Make sure that you include the correct number of times (f) the value (X) appears in the set:
3, 4, 4, 5, 5, 6, 6, 7, 7, 7, 7, 8 (Notice that the total number of scores, that is the sum of the f’s, is 12)
2. To compute the mean, add up all the values and then divide the total by the number of scores in the set:
69
12 = 5.75
3. To compute the median, find the physical center of all of the values once they are put in order:
3, 4, 4, 5, 5, 6 * 6, 7, 7, 7, 7, 8 The physical center * would have ½ of all the values below it and ½ of all the values above it. So, if we have a total of 12 values, 6 of the values would fall below * and 6 of the values would fall above *. There are 2 values of “6” in this data set and the * falls right in the middle of the 2 values of “6.” Therefore, the value of the median is “6.”
4. To compute the mode, find the X value with the highest frequency (f): Here, that is “7” because 7 falls 4 times in this data set, more frequently than any other value in the set.
Problem 2:
a. Compute the variance and standard deviation for the following data:
i. 3, 4, 4, 5, 5, 6, 6, 7, 7, 7, 7, 8 These are data are for a population.
b. Compute the variance and standard deviation for the same data, except now consider that the data are for a sample from a population.
(See pp. 109 and 116 for directions in textbook)
Solution:
Variance = X-M)2 /N; SD = (that is, SD = the square root of the Variance)
Set up your work to do the steps in order
X Values: 3, 4, 4, 5, 5, 6, 6, 7, 7, 7, 7, 8 (a population)
|
X value |
X - M |
(X-M)2 |
(Population variance) |
Population SD) |
V (Sample Variance) |
SD (Sample SD) |
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3 |
3-5.75 = - 2.75 |
-2.75 * -2.75 = 7.5625 |
Sum the squares of the deviations: (X-M)2 = 26.25 and divide by N:
(X-M)2/N = 26.25/12 = 2.19
2 = 2.19 |
= 1.48 |
Sum the squares of the deviations: (X-M)2 = 26.25 and divide by N - 1:
(X-M)2/N - 1 = 26.25/11 = 2.39
V = 2.38 |
SD
SD
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4 |
4-5.75 = - 1.75 |
-1.75 * -1.75 = 3.0625 |
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4 |
4-5.75 = - 1.75 |
3.0625 |
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5 |
5-5.75 = - .75 |
.5625 |
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5 |
5-5.75 = - .75 |
.5625 |
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6 |
6-5.75 = + .25 |
.0625 |
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6 |
6-5.75 = + .25 |
.0625 |
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7 |
7-5.75 = + 1.25 |
1.5625 |
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7 |
7-5.75 = + 1.25 |
1.5625 |
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7 |
7-5.75 = + 1.25 |
1.5625 |
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7 |
7-5.75 = + 1.25 |
1.5625 |
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8 |
8-5.75 = + 2.25 |
5.0625 |
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Compute the Mean (M) of X |
If all your deviations from the mean are correct, they should add up to 0. -7.75 + 7.75 = 0 |
Square each value you have in the preceding column (* means the same thing as X for multiply) |
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X =69 ; N = 12 |
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X/N = 5.75 |
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M = 5.75 |
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Problem 3:
Four students each received a raw score of 22 on a different test. Compute the z-score for the raw score value of 22 for each student, given information about the distributions of X values for each test. Which student (a, b, c, or d) had the highest score value, compared to the normative group who took the same test? Which student did the poorest in this group, compared to the normative group who took the same test? Why?
a. X = 22, when M = 20, SD = 5, Z = ?
b. X = 22, when M = 30, SD = 5, Z = ?
c. X = 22, when M = 16, SD = 2, Z = ?
d. X = 22, when M = 10, SD = 8, Z = ?
(See p. 160 in textbook)
Solution: X – M
Z = SD
a. 22 - 20 2
Z = 5 = 5 = .4 (Notice that the Z-value is positive because X is greater than M)
b. 22 - 30 - 8
Z = 5 = 5 = - 1.6 (Notice that the Z-value is negative because X is less than M)
c. 22 - 16 6
Z = 2 = 2 = 3 (Notice that the value is positive because X is greater than M)
d. 22 - 10 12
Z = 8 = 8 = 1.5 (Notice that the value is positive because X is greater than M)
Student c had the highest score value, compared to others who took the same test, because his score was 3 SD units above the mean, which would be an extremely high.
Student b did the poorest in this group, compared to others who took the same test, because his score was 1.6 SD units below the mean.
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