运用SPSS进行信度分析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）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。

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。

CorrelationsDescriptive Statistics11.8000 4.661901012.8000 5.1380910OCTAPRMean Std. Deviation N纸笔计算结果：N=1097.97228.557521.2104.151172)8744.4)(4226.4()8.12)(8.11()10/1720(==-=-=XX r复本信度（Alternate-form Reliability ）Correlations纸笔计算结果：N=1096.9627.5883798.572.1531.159)2891.2)(4413.2()6.12)(2.12()10/1591(==-=-=xx r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BivariateCorrelationsDescriptive Statistics1.4000.5477252.2000.836665ODDS EVENSMean Std. Deviation N纸笔计算结果Ⅰ. 计算两个”半测验”的相关 N=587.8729.3665922.08.34.3)7483)(.4899(.)2.2)(4.1()5/17(==-=-=XX rⅡ 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)))(1(1))((XX XX XX current N current N NEW γγγ-+=93.9321.8729.17458.1)8729)(.12(1)8729.0)(2(==+=-+=XX NEW γGuttman prophesy formula 哥德曼校正公式 (当两个半测验变异数不等时使用))1(2222XE O OE r σσσ+-= 888.0)8.17.03.01(2))3416.1()8367.0()5477.0(1(2222=+-=+-=OENEW γ*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 CoefficientsN of Cases = 5.0 N of Items = 6Correlation 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 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 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 = .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 2Alpha for part 1 = -.9000 Alpha for part 2 = .6923內部一致性（Internal Consistency Coefficient）【计算α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person Item 1 Item 2 Item 3 Item 4 Item 5Joe 3 4 4 3 5Sam 4 3 4 3 3Sue 2 3 3 2 3步骤一步骤二按【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。

出现下列对话框候选取下列勾选后按键步骤三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_1 ITEM_2 ITEM_3 ITEM_4 ITEM_5 ITEM_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 StatisticsScale Scale CorrectedMean Variance Item- Squared Alpha if Item if Item Total Multiple if Item Deleted Deleted Correlation Correlation DeletedITEM_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 .8333Reliability Coefficients 5 itemsAlpha = .8457 Standardized item alpha = .8609纸笔计算结果σi =.4722 .6667 .4722 .2222 .5833X =16.1667, σ2=7.4722 σ=2.7335 k=5 N=685.84575.)3234.1)(25.1()4722.74166.21)(25.1()4722.75833.2222.4722.6667.4722.1(45==-=-=++++-=α步骤一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 MatrixITEM_1 ITEM_2 ITEM_3 ITEM_4 ITEM_5 ITEM_6ITEM_1 1.0000ITEM_2 -.6667 1.0000ITEM_3 .4082 .4082 1.0000ITEM_4 1.0000 -.6667 .4082 1.0000ITEM_5 -.6124 .4082 -.2500 -.6124 1.0000ITEM_6 .1667 .1667 .4082 .1667 .4082 1.0000 * * * Warning * * * Determinant of matrix is close to zero: 8.426E-36Statistics based on inverse matrix for scale ALPHAare meaningless and printed as .N of Cases = 5.0Item-total StatisticsScale Scale CorrectedMean Variance Item- Squared Alphaif Item if Item Total Multiple if Item Deleted Deleted Correlation Correlation DeletedITEM_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 . -.1042R 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 itemsAlpha = .3273 Standardized item alpha = .3307纸笔计算结果Pi .6 .6 .2 .6 .2 .6 qi .4 .4 .8 .4 .8 .4 (p)(q).24.24 .16 .24.16.24X =2.8, σ2=1.76 σ=1.3266 k=6 N=533.3273.)2727)(.2.1()76.148.)(2.1()76.128.1()76.1((5620====-=-KR。