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PSY 216
Assignment 4 Answers
1. Problem 2 from the text
Can SS ever have a value less than zero? Explain your answer.
SS = Σ(X - )2. When you square a real number, the result is always non-negative. The sum of non-negative
numbers must be non-negative.
2. Problem 4 from the text
What does it mean for a sample to have a standard deviation of zero? Describe the scores in such a sample.
If the standard deviation is 0 then the variance is 0 and the mean of the squared deviation scores must be 0.
The sum of the squared deviation scores must be equal to 0 for the mean of the squared deviation scores to be
equal to 0. Since the squared deviation scores must be non-negative because they are squared, all of the
squared deviation scores must be 0; otherwise, the sum would be non-zero. The only way that each squared
deviation score can be equal to 0 is if all of the scores equal the mean. Thus, when the standard deviation
equals 0, all the scores are identical and equal to the mean.
3. Problem 5 from the text
Explain why the formulas for sample variance and population variance are different.
The extreme scores in a population that is approximately normal in shape are less likely to be
included in a sample because of their rarity. Thus, the sample is usually less variable than the
population. By dividing the sum of squares of the sample by a slightly smaller number (n – 1 instead
of n), the variance will be slightly larger, correcting for underestimation.
4. Problem 7 from the text
On an exam with a mean of M = 78, you obtain a score of X = 84.
a. Would you prefer a standard deviation of s = 2 or s = 10? (Hint: Sketch each distribution
and find the location of your score.)
One should prefer s = 2 as a score of 84 is an extremely large value in such a distribution.
You likely did much better than everyone else. With s = 10, your score is less extreme and
others likely did better than you.
b. If your score was X = 72, would you prefer s = 2 or s = 10. Explain your answer.
One should prefer s = 10 as your below average performance is not an extreme score. With
s = 2, few, if any, people performed less well than you on the exam.
5. Problem 8 from the text
A population has a mean of μ = 30 and a standard deviation of σ = 5. (Problem 8 from the text)
a. If 5 points were added to every score in the population, what would be the new values for
the mean and standard deviation?
Adding a constant to every score increases the mean by the same constant amount. Thus, μ
= 30 + 5 = 35.
Adding a constant to every score has no effect on the standard deviation. σ = 5
b. If every score in the population were multiplied by 3 what would be the new values for the
mean and standard deviation?
Multiplying each score by a constant also multiplies the mean by the same constant. Thus,
μ = 30 * 3 = 90. In a population, multiplying each score by a constant also multiplies the
standard deviation by the constant. Thus, σ = 5 * 3 = 15
6. Problem 13 from the text
Calculate the mean and SS (sum of squared deviations) for each of the following samples. Based on
the value for the mean, you should be able to decide which SS formula is better to use.
a. Sample A: 1, 4, 8, 5
M = ΣX / n = (1 + 4 + 8 + 5) / 4 = 18 / 4 = 4.5
SS = (ΣX2 – (ΣX)2/ N)
= (12 + 42 + 82 + 52 – (1 + 4 + 8 + 5)2 / 4)
= (1 + 16 + 64 + 25 – 182 / 4)
= (106 – 324 / 4)
= (106 – 81)
= 25
Less preferred method because the mean is not an integer:
SS = Σ(X – M)2
= (1 – 4.5)2 + (4 – 4.5)2 + (8 – 4.5)2 + (5 – 4.5)2
= -3.52 + -0.52 + 3.52 + 0.52
=12.25 + 0.25 + 12.25 + 0.25
= 25
b. Sample B: 3, 0, 9, 4
M = ΣX / n = (3 + 0 + 9 + 4) / 4 = 16 / 4 = 4
SS = Σ(X – M)2
= (3 – 4)2 + (0 – 4)2 + (9 – 4)2 + (4 – 4)2
= -12 + -42 + 52 + 02
=1 + 16 + 25 + 0
= 42
SS = (ΣX2 – (ΣX)2/ N)
= (32 + 02 + 92 + 42 – (3 + 0 + 9 + 4)2 / 4)
= (9 + 0 + 81 + 16 – 162 / 4)
= (106 – 256 / 4)
= (106 – 64)
= 42
7. Problem 15 from the text
For the data in the following sample: 8, 1, 5, 1, 5
a. Find the mean and standard deviation.
M = ΣX / N = (8 + 1 + 5 + 1 + 5) / 5 = 20 / 5 = 4
s = √(SS / df)
= √(Σ(X - M)2/(N – 1)
= √((8 - 4)2 + (1 – 4)2 + (5 – 4)2 + (1 – 4)2 + (5 – 4)2) / (5 – 1)
= √(16 + 9 + 1 + 9 + 1) / 4
= √36/4
=3
b. Now change the score of X = 8 to X = 18 and find the new mean and standard deviation.
18, 1, 5, 1, 5
M = ΣX / N = (18 + 1 + 5 + 1 + 5) / 5 = 30 / 5 = 6
s = √(SS / df)
= √(Σ(X - M)2/(N – 1)
= √((18 - 6)2 + (1 – 6)2 + (5 –6)2 + (1 – 6)2 + (5 – 6)2) / (5 – 1)
= √(144 + 25 + 1 +25 + 1) / 4
= √196/4
=7
c. Describe how one extreme score influences the mean and standard deviation.
The mean increased when a smaller score was replaced by a larger score.
The standard deviation increased when a score close to the mean was replaced by a score farther away
from the mean.
8. Problem 18 from the text
Calculate SS, variance, and standard deviation for the following population of N = 7 scores: 8, 1, 4, 3, 5, 3, 4.
(Note: The definitional formula works well with these scores.)
M = ΣX / N = (8 + 1 + 4 + 3 + 5 + 3 + 4) / 7 = 28 / 7 = 4
SS = Σ(X – M)2
= (8 – 4)2 + (1 – 4)2 + (4 – 4)2 + (3 – 4)2 + (5 – 4)2 + (3 – 4)2 + (4 – 4)2
= 42 + -32 + 02 + -12 + 12 + -12 + 02
= 16 + 9 + 0 + 1 + 1 + 1 + 0
= 28
σ2 = SS / N
= 28 / 7
=4
σ = √ σ2
= √4
=2
9. Use SPSS to graph a histogram of the number of younger siblings variable in the class data set. For
that variable, use SPSS to find the following statistics and report their values in the following table:
Statistic
Value
Mean
0.95
Median
1.00
Mode
0
Standard deviation
1.089
Variance
1.186
Interquartile range
Q3 – Q1 = 75th %tile –
25th %tile = 1.25 – 0 =
1.25
4
Range
Attach your SPSS output to your answers.
There a numerous ways of accomplishing this in SPSS. One way is to click on Analyze | Descriptive
Statistics | Frequencies. Move the Number of Younger Siblings variable into the Variables list. Click
on the Statistics button and select Mean, Median, Mode, Std. deviation, Variance, Range, and Quartiles
(for the interquatile range). Click Continue. Click the Charts button. Select Histograms. Click
Continue. Click OK.
Statistics
Number of younger siblings
N
Valid
Missing
Mean
38
0
.95
Median
1.00
Mode
0
Std. Deviation
1.089
Variance
1.186
Range
Percentiles
4
25
.00
50
1.00
75
1.25