运用SPSS进行信度分析报告

发布时间:2019-10-22 10:06:42

SPSS信度分析步骤

资料输入

Data输入页

变项定义页



信度分析

1.再测信度(Test-Retest Reliability

2.复本信度(Alternate-form Reliability

3.折半信度(Split-half Reliaility

4.內部一致性(Internal Consistency Coefficient)【计算α系数】

再测信度(Test-Retest Reliability

步骤一 按【Analyze】→【Correlate】→【Bivariate…

步骤二 会出现下面的对话框,将左边两变项选入右边「Variables」内,在「Correlation Coefficients」方盒内选取「□Pearson」;在「Test of Significance」方盒内选取「□Two-tailed」;勾选最下面的「□Flag significant correlations」,之后按键。

补充 若想呈现平均及标准差可在按键前按进入下个对话框,在Statistics的方盒内选取「□Means and standard deviations」,按继续。

Correlations

纸笔计算结果:

Person

A

B

C

D

E

F

G

H

I

J

σ

Oct

18

16

5

13

15

16

12

5

8

10

11.8

4.4226

Apr

18

18

6

16

17

16

14

5

7

11

12.8

4.8744

X1 X2

324

288

30

208

255

256

168

25

56

110

X1X2 =1720

N=10

复本信度(Alternate-form Reliability

Correlations

纸笔计算结果:

Person

A

B

C

D

E

F

G

H

I

J

σ

Form A

16

12

14

10

9

11

13

9

16

12

12.2

2.4413

Form B

15

12

15

10

10

12

14

9

16

13

12.6

2.289

XA XB

240

144

210

100

90

132

182

81

256

156

ΣXAXB=1591

N=10

折半信度(Split-half Reliability

步骤一 输入资料

步骤二 转换资料为数字

按【Transform】→【Recode】→【Into Same Variables…

出现下面的对话框后将左边方格内item1~item6选至右边String Variables内后点选

出现下列对话框后,将”N”定义为”0”,将”Y”定义为”1”后按

之后便会将资料转换成下面的数字

步骤三 string的属性改为numeric

步骤四 计算奇数题和偶数题的和

按【Transform】→【Compute…】即出现下面的对话框

结束后便会在spss Data Editor对话框中出现奇数题和偶数题的和

步骤四 执行Bivariate

Correlations

纸笔计算结果

. 计算两个”半测验”的相关

Student

Joe

Sam

Sue

Peg

Gil

σ

Odds

2

1

2

1

1

1.4

.4899

Evens

3

2

3

1

2

2.2

.7483

X1X2

6

2

6

1

2

X1X2 =17

N=5

校正相关系数为折半信度

Spearmen-Brown prophesy formula 史比校正公式 (当两个半测验变异数相等时使用)

Guttman prophesy formula 哥德曼校正公式 (当两个半测验变异数不等时使用)

*折半信度* 折半信度也可直接使用SPSS計算

步骤一 输入资料

步驟二 按【Analyze】→【Scale】→【Reliability Analysis】将左边方格内的变项依所需次序分前后半选入右边items的方格内,在左下角的Model框中选取Split-half后按键,再按

Reliability

****** Method 1 (space saver) will be used for this analysis ******

R E L I A B I L I T Y A N A L Y S I S - S C A L E (S P L I T)

Reliability Coefficients

N of Cases = 5.0 N of Items = 6

Correlation between forms = .8729 Equal-length Spearman-Brown = .9321

Guttman Split-half = .8889 Unequal-length Spearman-Brown = .9321

3 Items in part 1 3 Items in part 2

Alpha for part 1 = -2.5000 Alpha for part 2 = .0000

透过平均值可看出其难度

平均高的难度低

平均低的难度高

排序由难度低到高2 6 1 4 5 3,在丢入变项时依单偶分为:2 1 56 4 3两组,排列数据时前后排列。

Correlations

Correlations

****** Method 1 (space saver) will be used for this analysis ******

R E L I A B I L I T Y A N A L Y S I S - S C A L E (S P L I T)

Reliability Coefficients

N of Cases = 5.0 N of Items = 6

Correlation between forms = .0000 Equal-length Spearman-Brown = .0000

Guttman Split-half = .0000 Unequal-length Spearman-Brown = .0000

>Note # 11999

>The correlation between forms (halves) of the test is negative. This

>violates reliability model assumptions. Statistics which are functions of

>this value may have estimates outside theoretically possible ranges.

3 Items in part 1 3 Items in part 2

Alpha for part 1 = -.9000 Alpha for part 2 = .6923

內部一致性(Internal Consistency Coefficient)【计算α系数】

步骤一 输入资料

步骤二 按【Analyze】→【Scale】→【Reliability Analysis

将左边方格内的变项全选入右边items的方格内,在左下角的Model框中选取Alpha后按键。

步骤三 出现下列对话框候选取下列勾选后按

Reliability

****** Method 2 (covariance matrix) will be used for this analysis ******

R E L I A B I L I T Y A N A L Y S I S - S C A L E (A L P H A)

Correlation Matrix

ITEM_1 ITEM_2 ITEM_3 ITEM_4 ITEM_5

ITEM_1 1.0000

ITEM_2 .2970 1.0000

ITEM_3 .7647 .5941 1.0000

ITEM_4 .6860 .4330 .8575 1.0000

ITEM_5 .1588 .8018 .4763 .4629 1.0000

N of Cases = 6.0

Item-total Statistics

Scale Scale Corrected

Mean Variance Item- Squared Alpha

if Item if Item Total Multiple if Item

Deleted Deleted Correlation Correlation Deleted

ITEM_1 13.0000 6.4000 .5251 .6471 .8472

ITEM_2 13.1667 5.3667 .6757 .7500 .8116

ITEM_3 12.3333 5.4667 .8333 .8588 .7642

ITEM_4 13.5000 6.7000 .7481 .7857 .8093

ITEM_5 12.6667 5.8667 .5922 .7143 .8333

Reliability Coefficients 5 items

Alpha = .8457 Standardized item alpha = .8609

纸笔计算结果

Question

1

2

3

4

5

Total Score

Joe

3

4

4

3

5

19

Sam

4

3

4

3

3

17

Sue

2

3

3

2

3

13

Peg

4

4

5

3

4

20

Gil

3

2

4

3

3

15

Dot

3

2

3

2

3

13

σi2 =

.4722

.6667

.4722

.2222

.5833

=16.1667, σ2=7.4722 σ=2.7335 k=5 N=6

步骤一 转换资料Y正确为1N不正确为0

步骤二 与题四求内部ㄧ致性的步骤相同

Reliability

****** Method 2 (covariance matrix) will be used for this analysis ******

R E L I A B I L I T Y A N A L Y S I S - S C A L E (A L P H A)

Correlation Matrix

ITEM_1

ITEM_2

ITEM_3

ITEM_4

ITEM_5

ITEM_6

ITEM_1

1.0000

ITEM_2

-.6667

1.0000

ITEM_3

.4082

.4082

1.0000

ITEM_4

1.0000

-.6667

.4082

1.0000

ITEM_5

-.6124

.4082

-.2500

-.6124

1.0000

ITEM_6

.1667

.1667

.4082

.1667

.4082

1.0000

* * * Warning * * * Determinant of matrix is close to zero: 8.426E-36

Statistics based on inverse matrix for scale ALPHA

are meaningless and printed as .

N of Cases = 5.0

Item-total Statistics

Scale Scale Corrected

Mean Variance Item- Squared Alpha

if Item if Item Total Multiple if Item

Deleted Deleted Correlation Correlation Deleted

ITEM_1 2.2000 1.7000 .1400 . .2941

ITEM_2 2.2000 2.2000 -.1846 . .5114

ITEM_3 2.6000 1.3000 .6864 . -.0962

ITEM_4 2.2000 1.7000 .1400 . .2941

ITEM_5 2.6000 2.3000 -.2212 . .4891

ITEM_6 2.2000 1.2000 .5833 . -.1042

R E L I A B I L I T Y A N A L Y S I S - S C A L E (A L P H A)

Reliability Coefficients 6 items

Alpha = .3273 Standardized item alpha = .3307

纸笔计算结果

Question

1

2

3

4

5

6

Total Score

A

Y

Y

Y

Y

N

Y

5

B

Y

N

N

Y

N

Y

3

C

Y

N

N

Y

N

N

2

D

N

Y

N

N

N

N

1

E

N

Y

N

N

Y

Y

3

Pi

.6

.6

.2

.6

.2

.6

qi

.4

.4

.8

.4

.8

.4

(p)(q)

.24

.24

.16

.24

.16

.24

=2.8, σ2=1.76 σ=1.3266 k=6 N=5

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运用SPSS进行信度分析报告

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