运用SPSS进行信度分析报告

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名学生，以下是测验结果，请计算信度。PersonABCDEFGHIJOct18165131516125810Ap

2、r18186161716145711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 significantcorrelations，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 Meansand standarddeviations，按继续。Correlat

3、ionsDescriptive StatisticsOCTMeanStd. DeviationN11.80004.6619010APR12.80005.1380910CorrelationsOCTPearson CorrelationSig. (2-tailed)NAPRPearson CorrelationSig. (2-tailed)NOCTAPR1.972*.0001010.972*1.000.1010*. Correlation is significant at the 0.01 level (2-tailed).纸笔计算结果：PersonABCDEFGHIJXOct18165131

4、51612581011.8 4.4226Apr1818616171614571112.8 4.8744X1 X2324288302082552561682556110X1 X2 =1720N=10(1720 /10)(11.8)(12.8)172151.04.97r XX.97228( 4.4226)( 4.8744)21.557521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 A1612

5、14109111391612Form B1512151010121491613CorrelationsDescriptive StatisticsFORM_AMeanStd. DeviationN12.20002.5733710FORM_B12.60002.4129310CorrelationsFORM_APearson CorrelationSig. (2-tailed)NFORM_BPearson CorrelationSig. (2-tailed)NFORM_AFORM_B1.963*.0001010.963*1.000.1010*. Correlation is significant

6、 at the 0.01 level (2-tailed).纸笔计算结果：PersonABCDEFGHIJXForm A16121410911139161212.22.4413Form B151215101012149161312.62.289X XB2401442101009013218281256156XX =1591AA B.N=10r xx(1591/ 10)(12.2)(12.6)159.1153.72(2.4413)(2.2891).9627 .965.5883798折半信度（ Split-half Reliability）6 题的随堂测验施测 5 位学生， Y 表示答对， N 表

7、示答错，以下是测验结果，请计算信度。PersonItem 1Item 2Item 3Item 4Item 5Item 6JoeYYYYNYSamYNNYNYSueYYNYYYPegNYNNYNGilNYNNYY步骤一输入资料.步骤二转换资料为数字按【Transform 】【Recode】【IntoSameVariables 】.出现下面的对话框后将左边方格内item1item6 选至右边 StringVariables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

8、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.CorrelationsDescriptive StatisticsODDSMeanStd. DeviationN1.4000.547725EVENS2.2000.836665.CorrelationsODDSPearson CorrelationODDSEVENS1.873Sig. (2-tailed).053N55EVENSPearson Correlation.8731Sig. (2-tailed).

9、053.N55纸笔计算结果 . 计算两个”半测验”的相关StudentJoeSamSuePegGilXOdds212111.4.4899Evens323122.2.7483X1X262612X1X2 =17N=5(17 / 5) (1.4)(2.2)3.4 3.08r XX.8729 .87(.4899)(.7483).3665922 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 )NEW XX( N )(current XX )( N 1)(current XX )1(2)(0.8729)1.7458NEW

10、 XX1)(. 8729).9321 .931 (21 .8729Guttman prophesy formula哥德曼校正公式 ( 当两个半测验变异数不等时使用 )22rOE2(1O2E )XNEW OE2(1(0.5477) 2(0.8367) 2)2(1 0.3 0.7 ) 0.888(1.3416) 21.8* 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 】【 ReliabilityAnalysis 】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

11、键，再按。Reliability* Method 1 (space saver) will be used for this analysis *RELIABILITY ANALYSIS - SCALE (SPLIT)Reliability CoefficientsN of Cases = 5.0N of Items = 6Correlation between forms = .8729 Equal-length Spearman-Brown =.9321Guttman Split-half =.8889Unequal-length Spearman-Brown =.93213 Items

12、in part 13 Items in part 2Alpha for part 1 =-2.5000Alpha for part 2 =.0000.Descriptive StatisticsNSumMeanITEM_153.60ITEM_351.20ITEM_553.60ITEM_254.80ITEM_453.60ITEM_654.80Valid N(listwise)5CorrelationsCorrelationsSPearsonSF1.000CorrelationSig. (2-tailed).1.000N55FPearson.0001CorrelationSig. (2-taile

13、d)1.000.N55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。Correlations* Method 1 (space saver) will be used for this analysis *RELIABILITY ANALYSIS - SCALE (SPLIT)Reliability CoefficientsN of Cases = 5.0N of Items = 6Correlation between forms = .0000 Equal-length

14、 Spearman-Brown =.0000Guttman Split-half =.0000Unequal-length Spearman-Brown =.0000Note # 11999The correlation between forms (halves) of the test is negative. Thisviolates reliability model assumptions. Statistics which are functions ofthis value may have estimates outside theoretically possible ran

15、ges.3 Items in part 13 Items in part 2Alpha for part 1 =-.9000Alpha for part 2 =.6923.內部一致性（ Internal Consistency Coefficient）【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 】【 Rel

16、iability 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 be used for this analysis *RELIABILITYANALYSIS-SCALE(ALPHA)Correlation MatrixITEM_1ITEM_2ITEM_3ITEM_4ITEM_5ITEM_11.0000ITEM_2.29701.0000ITEM_3.7647.59411.0000IT

17、EM_4.6860.4330.85751.0000ITEM_5.1588.8018.4763.46291.0000N of Cases =6.0Item-total Statistics.ScaleMeanScale VarianceCorrectedItem-.SquaredAlpha.if Itemif ItemTotalMultipleif ItemDeletedDeletedCorrelationCorrelationDeletedITEM_113.00006.4000.5251.6471.8472ITEM_213.16675.3667.6757.7500.8116ITEM_312.3

18、3335.4667.8333.8588.7642ITEM_413.50006.7000.7481.7857.8093ITEM_512.66675.8667.5922.7143.8333Reliability Coefficients5 itemsAlpha = .8457Standardized item alpha =.8609纸笔计算结果Question12345Total ScoreJoe3443519Sam4343317Sue2332313Peg4453420Gil3243315Dot3232313i 2 = .4722 .6667 .4722 .2222 .5833X =16.166

19、7,2=7.4722 =2.7335 k=5 N=65 (1.4722.6667 .4722 .2222 .5833)(1.25)(12.4166 ) (1.25)(1 .3234) .84575 .8547.47227.4722有 6 题的随测验施测 5 名学生， Y 表示答对， N 表示答错，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

20、 used for this analysis *RELIABILITYANALYSIS-SCALE(ALPHA)ITEM_1Correlation Matrix ITEM_2 ITEM_3ITEM_4ITEM_5ITEM_6ITEM_11.0000ITEM_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 m

21、atrix is close to zero: 8.426E-36 Statistics based on inverse matrix for scale ALPHAare meaningless and printed as .N of Cases =5.0Item-total StatisticsScaleMeanif ItemDeletedScale Variance if Item DeletedCorrectedItem-SquaredAlphaTotalMultipleif ItemCorrelationCorrelationDeletedITEM_12.20001.7000.1400.2941ITEM_22.20002.2000-.1846.5114.ITEM_32.60001.3000.6864.-.0962I