运用SPSS进行信度分析

1、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）【计算系数】再测信度（Test-Retest Reliability）某空间性向测验有20题单选题，分别在十月与第二年四月施测同一组10名学生，以下是测验结果，请计算信度。PersonABCDEFGHIJOct18165131516125810Apr1818616171

2、6145711 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：PersonAB

3、CDEFGHIJOct1816513151612581011.84.4226Apr1818616171614571112.84.8744X1 X2324288302082552561682556110X1X2 =1720N=10复本信度（Alternate-form Reliability）某自我概念量表，有20题是非题，题本A与B 分别施测同一组10个人，分数愈高表示愈具有正向自我概念，以下是施测结果，请计算信度。PersonABCDEFGHIJForm A161214109111391612Form B1512151010121491613Correlations纸笔计算结果：Person

4、ABCDEFGHIJForm A16121410911139161212.22.4413Form B151215101012149161312.62.289XA XB2401442101009013218281256156XAXB=1591N=106题的随堂测验施测5位学生， Y表示答对，N表示答错，以下是测验结果，请计算信度。PersonItem 1Item 2Item 3Item 4Item 5Item 6JoeYYYYNYSamYNNYNYSueYYNYYYPegNYNNYNGilNYNNYY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

5、【Transform】【Recode】【Into Same Variables】出现下面的对话框后将左边方格内item1item6选至右边String Variables内后点选键出现下列对话框后，将”N”定义为”0”，将”Y”定义为”1”后按键之后便会将资料转换成下面的数字步骤三 将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BivariateCorrelations纸笔计算结果. 计算两个”半测验”的相关StudentJoe

6、SamSuePegGilOdds212111.4.4899Evens323122.2.7483X1X262612X1X2 =17N=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

7、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 CoefficientsN of Cases = 5.0 N of Items = 6Correlation between forms = .8729 Equal-length Spearman-Brown = .9321Guttman Split-half = .888

8、9 Unequal-length Spearman-Brown = .9321 3 Items in part 1 3 Items in part 2Alpha 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 5、6 4 3两组，排列数据时前后排列。CorrelationsCorrelations* Method 1 (space saver) will be used for this analysis * R E L I A

9、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 CoefficientsN of Cases = 5.0 N of Items = 6Correlation between forms = .0000 Equal-length Spearman-Brown = .0000Guttman Split-half = .0000 Unequal-length Spearman-Brown = .0000>Note # 11999>The correlation between forms (halves) of

10、 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 2Alpha for part 1 = -.9000 Alpha for part 2 = .6923內部一致性（Internal Consistency Coefficient

11、）【计算系数】有5题问答题的随测验施测5名学生，每题问答题配分是5分，以下是施测结果，请计算信度personItem 1Item 2Item 3Item 4Item 5Joe34435Sam43433Sue23323Peg44534Gil32433Dot32323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键按。Reliability* Method 2 (covariance matrix) will

12、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_5ITEM_1 1.0000ITEM_2 .2970 1.0000ITEM_3 .7647 .5941 1.0000ITEM_4 .6860 .4330 .8575 1.0000ITEM_5 .1588 .8018 .4763 .4629 1.0000 N of Cases = 6.0Item-total Stat

13、istics Scale Scale Corrected Mean Variance Item- Squared Alpha if Item if Item Total Multiple if Item Deleted Deleted Correlation Correlation DeletedITEM_1 13.0000 6.4000 .5251 .6471 .8472ITEM_2 13.1667 5.3667 .6757 .7500 .8116ITEM_3 12.3333 5.4667 .8333 .8588 .7642ITEM_4 13.5000 6.7000 .7481 .7857

14、.8093ITEM_5 12.6667 5.8667 .5922 .7143 .8333Reliability Coefficients 5 itemsAlpha = .8457 Standardized item alpha = .8609纸笔计算结果Question12345Total ScoreJoe3443519Sam4343317Sue2332313Peg4453420Gil3243315Dot3232313i2 =.4722.6667.4722.2222.5833=16.1667, 2=7.4722 =2.7335 k=5 N=6有6题的随测验施测5名学生，Y表示答对，N表示答错，

15、以下是施测结果，请计算信度。Question123456AYYYYNYBYNNYNYCYNNYNNDNYNNNNENYNNYY步骤一 转换资料Y正确为1，N不正确为0步骤二 与题四求内部致性的步骤相同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 MatrixITEM_1ITEM_2 ITEM_3ITEM_4ITEM_5ITEM_6ITEM_11.

16、0000ITEM_2-.66671.0000ITEM_3.4082.40821.0000ITEM_41.0000-.6667.40821.0000ITEM_5-.6124.4082-.2500-.61241.0000ITEM_6.1667.1667 .4082.1667.40821.0000* * * Warning * * * Determinant of matrix is close to zero: 8.426E-36 Statistics based on inverse matrix for scale ALPHA are meaningless and printed as .

17、N of Cases = 5.0Item-total Statistics Scale Scale Corrected Mean Variance Item- Squared Alpha if Item if Item Total Multiple if Item Deleted Deleted Correlation Correlation DeletedITEM_1 2.2000 1.7000 .1400 . .2941ITEM_2 2.2000 2.2000 -.1846 . .5114ITEM_3 2.6000 1.3000 .6864 . -.0962ITEM_4 2.2000 1.7000 .1400 . .2941ITEM_5 2.6000 2.3000 -.2212 . .4891ITEM_6 2.2000 1.2000 .5833 . -.1042 R E L I A B I L I