# S O C An urban planning commissioner has created a database describing a sample of the neighborhoods in her city and has developed a scale by which each area can be rated for the “quality of life" (this includes measures of pollution, noise, open space, services available, and so on). She has also asked samples of residents of these areas about their level of satisfaction with their neighborhoods. Is there statistically significant agreement between the commissioner’s objective ratings of quality of life and the respondents’ self-reports of satisfaction? Write a sentence or two explaining your findings, including the results of the chi square test and pattern of column percentages. Satisfaction Quality of life Low Moderate High Totals Low 21 15 6 42 Moderate 12 25 21 58 High Totals 8 41 17 57 32 59 57 157

### Essentials Of Statistics

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

Chapter
Section

### Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836
Chapter 10, Problem 10.12P
Textbook Problem
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## S O C An urban planning commissioner has created a database describing a sample of the neighborhoods in her city and has developed a scale by which each area can be rated for the “quality of life" (this includes measures of pollution, noise, open space, services available, and so on). She has also asked samples of residents of these areas about their level of satisfaction with their neighborhoods. Is there statistically significant agreement between the commissioner’s objective ratings of quality of life and the respondents’ self-reports of satisfaction? Write a sentence or two explaining your findings, including the results of the chi square test and pattern of column percentages. Satisfaction Quality of life Low Moderate High Totals Low 21 15 6 42 Moderate 12 25 21 58 HighTotals 8 41 17 57 32 59 57 157

To determine

To find:

The chi square value for the given information and test for its significance.

### Explanation of Solution

Given:

The given statement is,

An urban planning commissioner has created a database describing a sample of the neighborhoods in her city and has developed a scale by which each area can be rated for the “quality of life" (this includes measures of pollution, noise, open space, services available, and so on). She has also asked samples of residents of these areas about their level of satisfaction with their neighborhoods.

The given table of information is,

 Satisfaction Quality of Life Low Moderate High Totals Low 21 15 6 42 Moderate 12 25 21 58 High 8 17 32 57 Totals 41 57 59 157

Approach:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

The five step model for hypothesis testing is,

Step 1. Making assumptions and meeting test requirements.

Step 2. Stating the null hypothesis.

Step 3. Selecting the sampling distribution and establishing the critical region.

Step 4. Computing test statistics.

Step 5. Making a decision and interpreting the results of the test.

Formula used:

For a chi square, the expected frequency fe is given as,

fe=Rowmarginal×ColumnmarginalN

Where N is the total of frequencies.

The chi square statistic is given by,

χ2(obtained)=(fofe)2fe

Where fo is the observed frequency,

And fe is the expected frequency.

The degrees of freedom for the bivariate table is given as,

df=(r1)(c1)

Where r is the number of rows and c is the number of columns.

Calculation:

From the given information,

The observed frequency is given as,

fe=Rowmarginal×ColumnmarginalN……(1)

Substitute 42 for row marginal, 41 for column marginal and 157 for N in equation (1).

f1=42×41157=10.97

Substitute 42 for row marginal, 57 for column marginal and 157 for N in equation (1).

f2=42×57157=15.25

Substitute 42 for row marginal, 59 for column marginal and 157 for N in equation (1).

f3=42×59157=15.78

Substitute 58 for row marginal, 41 for column marginal and 157 for N in equation (1).

f4=58×41157=15.15

Substitute 58 for row marginal, 57 for column marginal and 157 for N in equation (1).

f5=58×57157=21.06

Substitute 58 for row marginal, 59 for column marginal and 157 for N in equation (1).

f6=58×59157=21.80

Substitute 57 for row marginal, 41 for column marginal and 157 for N in equation (1).

f7=57×41157=14.89

Substitute 57 for row marginal, 57 for column marginal and 157 for N in equation (1).

f8=57×57157=20.69

Substitute 57 for row marginal, 59 for column marginal and 157 for N in equation (1).

f9=57×59157=21.42

Consider the following table,

 fo fe fo−fe (fo−fe)2 (fo−fe)2/fe 21 10

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