spss homework chapter seven uchechi okani 07/29/2011 practice exercise...

SPSS HOMEWORK CHAPTER SEVEN

UCHECHI OKANI

07/29/2011

Practice Exercise 7.1

Use data set 2 in appendix B. In the population from which the sample was drawn, 20% of the

employees are clerical, 50% are technical, and 30% are professional. Determine whether or not

the sample drawn conforms to these values. Hint: you will need to customize expected

probabilities and enter the category values and relative percentages (2, 5, and 3). Also be sure

you have entered the variable category as nominal.

A new data set was created (CLASSIFY.sav) to represent the percentages of real population job

classification (CLASSIFY1) and the percentages of the participants’ job classification

(CLASSIFY2)

According to data from Appendix B data set 2, 4 out of 12 are clerical; 3 out of 12 are technical,

and 5 out of 12 are professional. In the population from which the sample was drawn, 20% of the

employees are clerical, 50% are technical, and 30% are professional.

2 variables were created CLASSIFY1 (population) and CLASSIFY2 (sample from Appendix B

data set 2). 10 rows of data for population representing proper percentages and 12 rows of data

from sample in Appendix B data set 2. Job classifications clerical, Technical, and professional

were coded 1, 2, and 3 respectively.

Data VIEW:

CLASSIFY1 CLASSIFY2

CLERICAL CLERICAL

CLERICAL CLERICAL

TECHNICAL CLERICAL

TECHNICAL CLERICAL

TECHNICAL TECHNICAL

TECHNICAL TECHNICAL

TECHNICAL TECHNICAL

PROFESSIONAL PROFESSIONAL



PROFESSIONAL

PROFESSIONAL

Chi-square test command was run using customize expected probabilities with the category values (1, 2,

3) and relative percentages (2, 5, and 3). Variable category was entered as nominal.

OUTPUT:

The output above states to “Retain the null hypothesis” that the categories of classify 2 occur with the

specified probabilities. This means that the job classifications in the sample (CLASSIFY2) occur with the

probabilities of the real population

The chi-square test is not significant

Result statement:

A chi-square goodness of fit test was calculated comparing the frequency of occurrence of 3

different job classifications in a sample (CLASSIFY2) with that of the population

(CLASSIFY1). It was hypothesized that each job classification would occur at equal number of

times with that of the population (CLASSIFY1). No significant deviation from the hypothesized

values was found (x2(2) = .50, p > .05)

(See Model view below)

Practice exercise 7.2

A researcher wants to know if individuals are more likely to help in an emergency when they are indoors

or when they are outdoors. Of 28 participants who were outdoors, 19 helped and 9 did not. Of 23

participants who were indoors, 8 helped and 15 did not. Enter these data, and find out if helping

behavior is affected by the environment. The key to this problem is in the data entry. (Hint: How many

participants were there, and what do you know about each participant)

In running the Chi-Square Test of Independence, the Cells option was used to display the percentages

of each variable especially because the groups are different sizes.

Data view:

INDOORS OUTDOORS

HELPED HELPED

HELPED HELPED

HELPED HELPED

HELPED HELPED

HELPED HELPED

HELPED HELPED

HELPED HELPED

HELPED HELPED

NOHELP HELPED

NOHELP HELPED

NOHELP HELPED

NOHELP HELPED

NOHELP HELPED

NOHELP HELPED

NOHELP HELPED

NOHELP HELPED

NOHELP HELPED

NOHELP HELPED

NOHELP HELPED

NOHELP NOHELP

NOHELP NOHELP

NOHELP NOHELP

NOHELP NOHELP

NOHELP

NOHELP

NOHELP

NOHELP

NOHELP

OUTPUT:

INDOORS * OUTDOORS Crosstabulation

OUTDOORS

Total HELPED NOHELP

INDOORS HELPED Count 8 0 8

Expected Count 6.6 1.4 8.0

% within INDOORS 100.0% .0% 100.0%

% within OUTDOORS 42.1% .0% 34.8%

% of Total 34.8% .0% 34.8%

NOHELP Count 11 4 15


% within INDOORS 73.3% 26.7% 100.0%

% within OUTDOORS 57.9% 100.0% 65.2%

% of Total 47.8% 17.4% 65.2%

Total Count 19 4 23


% within INDOORS 82.6% 17.4% 100.0%

% within OUTDOORS 100.0% 100.0% 100.0%

% of Total 82.6% 17.4% 100.0%

The first part of output above shows the counts and percentages.

Chi-Square Tests

Value df

Asymp. Sig. (2-

sided)

Exact Sig. (2-

sided)

Exact Sig. (1-

sided)

Pearson Chi-Square 2.582a 1 .108

Continuity Correctionb 1.060 1 .303

Likelihood Ratio 3.856 1 .050

Fisher's Exact Test .257 .154

Linear-by-Linear Association 2.470 1 .116

N of Valid Cases 23

Chi-Square Tests

Value df

Asymp. Sig. (2-

sided)

Exact Sig. (2-

sided)

Exact Sig. (1-

sided)

Pearson Chi-Square 2.582a 1 .108

Continuity Correctionb 1.060 1 .303

Likelihood Ratio 3.856 1 .050

Fisher's Exact Test .257 .154

Linear-by-Linear Association 2.470 1 .116

N of Valid Cases 23

a. 2 cells (50.0%) have expected count less than 5. The minimum expected count is 1.39.

b. Computed only for a 2x2 table

The second output section above shows the results of the chi-square test. The most commonly

used value is the Pearson chi-square, shown in the first row (value of 2.582).

This Chi-Square test is not significant and indicates that likelihood of individuals to help or not

help in an emergency is an independent event. This means that knowing how likely individuals

are to help or not help in an emergency when they are indoors does not tell us anything about

how likely individuals are to help or not help in an emergency when they are outdoors

Result statement:

A chi-square test of independence was calculated comparing the results of how individuals

INDOORS and OUTDOORS are likely to help in case of emergency. No significant relationship

was found (x2(1) = 2.582, p > .05)


Using practice data set 1 in Appendix B, determine if younger participants (

The output above states to retain the null hypothesis that the distribution of math SKILL scores

was the same for both age groups

The model viewer below shows relevant graphs and full information about the result

The participants who are old (26 and above) averaged 12.04 in their math-skills scores. The

participants who are young (25 and below) averaged 11.95 in their math-skills scores. The

second part of the output is the result of the Mann-Whitney U Test itself. The value was 66.50,

with a significance level of .975.

Result Statement:

A Mann-Whitney U Test was used to examine the difference in the math SKILLS of young (age

25 and under) and old (age 26 and above) participants. No significant difference in the math

SKILLS scores was found (U = 66.500, p > .05).


Use the RACE.sav data file to determine whether or not the outcome of short-distance races is

different from that of medium-distance races.

DARA VIEW:

Long medium short experience

1.00 4.00 6.00 2.00

2.00 3.00 4.00 2.00

3.00 2.00 7.00 2.00

4.00 5.00 3.00 2.00

5.00 1.00 10.00 1.00

6.00 8.00 5.00 1.00

7.00 7.00 12.00 1.00

8.00 6.00 1.00 1.00

OUTPUT:

The output above states to retain the null hypothesis that the outcome/results of short races was

not different from the outcome of medium races.

The output shows that no significant difference was found between the results of short-distance

and medium-distance races.

Result statement:

A Wilcox test examined the results of the shot-distance and medium-distance races. No

significant difference was found in the results (Z = .775, p > .05). Short-distance results were not

significantly different from medium-distance results.


Use the data in practice data set 3 in Appendix B. If anxiety is measured on an ordinal scale,

determine if anxiety levels changed over time. Phrase your results.

Output:

The output above states to reject the null hypothesis that there was no change over time in

anxiety levels. (This means there was significant change found over time in anxiety levels)

Result statement:

A Friedman test was conducted comparing the levels of anxiety over time (ANXPRE, ANX1,

and ANX4). A significant difference was found (x2(2) = 20.694, p < .05). There was significant

change in anxiety levels over time.

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