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    Strayer,Phoniex,
    Feb-1999 - Mar-2006

  • MBA.Graduate Psychology,PHD in HRM
    Strayer,Phoniex,University of California
    Feb-1999 - Mar-2006

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Category > Statistics Posted 13 Aug 2017 My Price 8.00

chi-square analysis.

Interpret the following chi-square analysis.

  1. What is the null hypothesis?
  2. Using an alpha level of .05, what do you conclude?
  3. If you reject the null hypothesis, which cells appear to account for the decision?

Crosstab

CAN PEOPLE BE TRUSTED

Total

CAN TRUST

CANNOT TRUST

DEPENDS

BOTH MEN AND WOMEN SHOULD CONTRIBUTE TO INCOME

Strongly Agree

Count

28

97

4

129

Expected Count

36.5

87.9

4.6

129.0

Std. Residual

-1.4

1.0

-.3

Agree

Count

65

208

11

284

Expected Count

80.3

193.5

10.2

284.0

Std. Residual

-1.7

1.0

.2

Neither Agree nor Disagree

Count

61

90

5

156

Expected Count

44.1

106.3

5.6

156.0

Std. Residual

2.5

-1.6

-.3

Disgree

Count

25

31

3

59

Expected Count

16.7

40.2

2.1

59.0

Std. Residual

2.0

-1.5

.6

Strongly Disagree

Count

2

10

0

12

Expected Count

3.4

8.2

.4

12.0

Std. Residual

-.8

.6

-.7

Total

Count

181

436

23

640

Expected Count

181.0

436.0

23.0

640.0

Value

df

Asymp. Sig. (2-sided)

Pearson Chi-Square

24.118a

8

.002

Likelihood Ratio

23.977

8

.002

Linear-by-Linear Association

9.333

1

.002

N of Valid Cases

640

a. 4 cells (26.7%) have expected count less than 5. The minimum expected count is .43.

Answers

(3)
Status NEW Posted 13 Aug 2017 02:08 PM My Price 8.00

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