   Chapter 4, Problem 4.4P

Chapter
Section
Textbook Problem

S O C In problem 3.5 at the end of Chapter 3, you calculated measures of central tendency for four variables for 15 respondents. Two of those variables are reproduced here. Calculate the mean (if necessary), the range, and the standard deviation for each variable. What information is added by the measures of dispersion? Write a paragraph summarizing these statistics. Respondent Age Attitude on Abortion (High score = Strong Opposition) A 18 10 B 20 9 C 21 8 D 30 10 E 25 7 F 26 7 G 19 9 H 29 6 I 31 10 J 55 5 K 32 4 L 28 3 M 23 2 N 24 1 O 32 9 3.5 S O C Data have been gathered on four variables for 15 respondents (see the following table). Find and report the appropriate measure of central tendency for each variable. (On the “Attitude on Abortion” scale, a high score indicates strong opposition.) Respondent Marital Status Racial/Ethnic Group Age Attitude on Abortion A Single White 18 10 B Single Hispanic 20 9 C Widowed White 21 8 D Married White 30 10 E Married Hispanic 25 7 F Married White 26 7 G Divorced Black 19 9 H Widowed White 29 6 I Divorced White 31 10 J Married Black 55 5 K Widowed Asian American 32 4 L Married Native American 28 3 M Divorced White 23 2 N Married White 24 1 O Divorced Black 32 9

To determine

To find:

The range and standard deviation of each given variable

Explanation

Given:

The following table gives the data on two variables for 15 respondents.

 Respondent Age Attitude on Abortion (High score = Strong Opposition) A 18 10 B 20 9 C 21 8 D 30 10 E 25 7 F 26 7 G 19 9 H 29 6 I 31 10 J 55 5 K 32 4 L 28 3 M 23 2 N 24 1 O 32 9

Formula used:

The formula to calculate range is given by,

Range=Highest valueLowest value

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 variable age - arrange the data in the increasing order.

The data in increasing order is given by,

18,19,20,21,23,24,25,26,28,29,30,31,32,32,55.

The highest age is 55 and the lowest age is 18.

The range is given by,

Range=Highest ageLowest age

Substitute 55 for highest age and 18 for lowest age in the above mentioned formula,

Range=5518=37

The size of the population is 15.

The mean is given by,

X¯=i=1NXiN

Substitute 15 for N, 18 for X1, 20 for X2 and so on in the above mentioned formula,

X¯=18+20+............+24+3215=41315=27.53 ……(1)

Consider the following table of sum of squares,

 Scores (Xi) (Xi−X¯) (Xi−X¯)2 18 −9.53 90.8209 20 −7.53 56.7009 21 −6.53 42.6409 30 2.47 6.1009 25 −2.53 6.4009 26 −1.53 2.3409 19 −8.53 72.7609 29 1.47 2.1609 31 3.47 12.0409 55 27.47 754.601 32 4.47 19.9809 28 0.47 0.2209 23 −4.53 20.5209 24 −3.53 12.4609 32 4.47 19.9809 ∑(Xi−X¯)=0 ∑(Xi−X¯)2=1119.734

From equation (1), substitute 18 for X1 and 27.53 for X¯ in (X1X¯).

(X1X¯)=(1827.53)(X1X¯)=9.53

Square the both sides of the equation.

(X1X¯)2=(9.53)2(X1X¯)2=90.8209

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=90.8209+56.7009+.............+12.4609+19.9809=1119.734 ……(2)

The standard deviation is given by,

s=(XiX¯)2N

From equation (2), substitute 1119

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