运用SPSS进行信度分析

发布时间:2020-04-04 07:39:44   来源:文档文库   
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运用SPSS进行信度分析


SPSS信度分析步骤

Data输入页

变项定义页


信度分析

1.信度(Test-Retest Reliability

2.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

信度(Test-Retest Reliability

步骤 按【Analyze】→【Correlate】→【Bivariate

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

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

word/media/image15.gif

Correlations

word/media/image17_1.png

word/media/image18_1.png

纸笔计算结果:

N=10

0c9797ef2becefabf0de0f678f990713.png

本信度(Alternate-form Reliability

Correlations

word/media/image22_1.png

word/media/image23_1.png

纸笔计算结果

N=10

76b1ba3f25ab6472d581e9aa0a59bc0d.png

折半信度(Split-half Reliability

步骤 输入资料

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 string的属性改为numeric

计算奇数题和偶数题的和

按【Transform】→【Compute】即出现下面的对话

结束后便会在spss Data Editor对话框中出现奇数题和偶数题的和

Bivariate

Correlations

word/media/image38_1.png

word/media/image39.gifword/media/image40_1.png

纸笔计算结果

. 计算两个”半测验”的相关

N=5

626058403c006bab676dfe06f8ce8976.png

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)

245a670776d0cefcd1c15d855224e7b0.png

4082648afd1695e3ed5a7e36dfb9a0b2.png

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

a2d4a85f74e7c3bf24dc2742f6ed316a.png

fd65ac13b16020696d8179daa6a388fc.png

*折半信度* 折半信度也可直接使用SPSS計算

步骤 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,再按

word/media/image48.gif

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

word/media/image50.gifCorrelation 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

Correlations

word/media/image52_1.png

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

word/media/image53.gifCorrelation 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

输入资料

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

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

word/media/image15.gif

步骤 出现下列对话框候选取下列勾选

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

word/media/image59.gif

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

1ce0cdaf66b9bf4b74d1f118c30080ef.png =16.1667, σ2=7.4722 σ=2.7335 k=5 N=6

3bdba21f3291dc3957521c0f1175acfd.png

步骤 转换资料Y正确为1N不正确为0

步骤 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

* * * 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

1ce0cdaf66b9bf4b74d1f118c30080ef.png=2.8, σ2=1.76 σ=1.3266 k=6 N=5

89eb84da52241d5f94d565c705df50d0.png

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