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Compute the standard deviation for both sets of data presented in problem 3.13 and reproduced here. Compare the standard deviation computed for freshmen with the standard deviation computed for seniors. What happened? Why? Does this change relate to what happened to the mean over the four-year period? How? What happened to the shapes of the underlying distribution? Freshmen 10 43 30 30 45 40 12 40 42 35 45 25 10 33 50 42 32 38 11 47 22 26 37 38 10 Seniors 10 45 35 27 50 35 10 50 40 30 40 10 10 37 10 40 15 30 20 43 23 25 30 40 10 3.13 S O C A sample of 25 freshmen at a major university completed a survey that measured their degree of racial prejudice (the higher the score, the greater the prejudice). a. Compute the median and mean scores for these data. 10 43 30 30 45 40 12 40 42 35 45 25 10 33 50 42 32 38 11 47 22 26 37 38 10 b. These same 25 students completed the same survey during the senior year. Compute the median and mean for this second set of scores, and compare them to the earlier set. What happened? 10 45 35 27 50 35 10 50 40 30 40 10 10 37 10 40 15 30 20 43 23 25 30 40 10

BuyFind

Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836
BuyFind

Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836

Solutions

Chapter
Section
Chapter 4, Problem 4.17P
Textbook Problem
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Compute the standard deviation for both sets of data presented in problem 3.13 and reproduced here. Compare the standard deviation computed for freshmen with the standard deviation computed for seniors. What happened? Why? Does this change relate to what happened to the mean over the four-year period? How? What happened to the shapes of the underlying distribution?

Freshmen
10 43 30 30 45
40 12 40 42 35
45 25 10 33 50
42 32 38 11 47
22 26 37 38 10
Seniors
10 45 35 27 50
35 10 50 40 30
40 10 10 37 10
40 15 30 20 43
23 25 30 40 10

3.13 S O C A sample of 25 freshmen at a major university completed a survey that measured their degree of racial prejudice (the higher the score, the greater the prejudice).

a. Compute the median and mean scores for these data.

10 43 30 30 45
40 12 40 42 35
45 25 10 33 50
42 32 38 11 47
22 26 37 38 10

b. These same 25 students completed the same survey during the senior year. Compute the median and mean for this second set of scores, and compare them to the earlier set. What happened?

10 45 35 27 50
35 10 50 40 30
40 10 10 37 10
40 15 30 20 43
23 25 30 40 10

Expert Solution
To determine

To find:

The standard deviation for freshmen and seniors and compare them.

Explanation of Solution

Given:

The following table shows the scores of 25 freshmen.

10 43 30 30 45
40 12 40 42 35
45 25 10 33 50
42 32 38 11 47
22 26 37 38 10

The following table shows the scores of 25 seniors.

10 45 35 27 50
35 10 50 40 30
40 10 10 37 10
40 15 30 20 43
23 25 30 40 10

Formula used:

Let the data values be Xi’s.

The formula to calculate mean is given by,

X¯=i=1NXiN

The formula to calculate standard deviation is given by,

s=(XiX¯)2N

Where, N is the population size.

Calculation:

For the freshmen,

The size of the population is 25.

The mean is given by,

X¯=i=1NXiN

Substitute 25 for N, 10 for X1, 40 for X2 and so on in the above mentioned formula,

X¯=10+40+...+1025=79325=31.72 ……(1)

Consider the following table of sum of squares,

(Xi) (XiX¯) (XiX¯)2
10 21.72 471.7584
40 8.28 68.5584
45 13.28 176.3584
42 10.28 105.6784
22 9.72 94.4784
43 11.28 127.2384
12 19.72 388.8784
25 6.72 45.1584
32 0.28 0.0784
26 5.72 32.7184
30 1.72 2.9584
40 8.28 68.5584
10 21.72 471.7584
38 6.28 39.4384
37 5.28 27.8784
30 1.72 2.9584
42 10.28 105.6784
33 1.28 1.6384
11 20.72 429.3184
38 6.28 39.4384
45 13.28 176.3584
35 3.28 10.7584
50 18.28 334.1584
47 15.28 233.4784
10 21.72 471.7584
(XiX¯)=0 (XiX¯)2=3927.04

From equation (1), substitute 10 for X1 and 31.72 for X¯ in (X1X¯).

(X1X¯)=(1031.72)(X1X¯)=21.72

Square the both sides of the equation.

(X1X¯)2=(21.72)2(X1X¯)2=471.7584

Proceed in the same manner to calculate (XiX¯)2 for all the 1iN for the rest data and refer table for the rest of the (XiX¯)2 values calculated. Then the value of (XiX¯)2 is calculated as,

(XiX¯)2=471.7584+68.5584+...+471.7584=3927.04 ……(2)

The standard deviation is given by,

s=(XiX¯)2N

From equation (2), substitute 3927.04 for (XiX¯)2 and 25 for N in the above mentioned formula,

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