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1、心理测验SPSS上机：信度分析数据输入Data输入页变项定义页Reliability Analysis本次课题包含：如何进行各种信度分析1. 再测信度（Test-Retest Reliability）2. 复本信度（Alternate-form Reliability）3. 折半信度（Split-half Reliability）4. 内部一致性（Internal Consistency Coefficient）【计算系数】5. KR 20【壹】 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，
2、请计算信度。PersonABCDEFGHIJOct18165131516125810Apr18186161716145711 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则可在按键前按进入下个对话框，在Statistics的方盒内选取Means
3、 and standard deviations，按继续。Correlations纸笔计算结果：PersonABCDEFGHIJOct1816513151612581011.84.4226Apr1818616171614571112.84.8744X1 X2324288302082552561682556110X1X2 =1720N=10【貳】 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 A16121410
4、9111391612Form B1512151010121491613做法与再测信度相同，请参照。Correlations纸笔计算结果：PersonABCDEFGHIJForm 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 6JoeYYYYNYSamYNNYNYS
5、ueYYNYYYPegNYNNYNGilNYNNYY【參】 折半信度（Split-half Reliability）步骤一 输入数据步骤二 转换数据为数字按【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”后按键之后便会将数据转换成下面的数字步骤三 将string的属性改为numeric步骤四 计算奇数题与偶数题的和按【Transform】【Compute】即出现下面的对话框（此为奇数题的范例，偶数题亦然）
6、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StudentJoeSamSuePegGilOdds212111.4.4899Evens323122.2.7483X1X262612X1X2 =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计算
7、 步骤一 输入数据步骤二 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 = 6
8、Correlation between forms = .8729 Equal-length Spearman-Brown = .9321Guttman 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标准的折半信度算法：先求各题的难度，再依难度加以排序，再施行上述的方法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
9、，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 = .
10、0000Guttman Split-half = .0000 Unequal-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 ranges. 3 Items in
11、 part 1 3 Items in part 2Alpha for part 1 = -.9000 Alpha for part 2 = .6923【肆】 内部一致性（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 输入数据步骤二 按【Analyze】【Scale】【Reliability Analysis
12、】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处，后按键按。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_5ITEM_1 1.0000ITEM_2 .2970 1.0000ITEM_
13、3 .7647 .5941 1.0000ITEM_4 .6860 .4330 .8575 1.0000ITEM_5 .1588 .8018 .4763 .4629 1.0000 N of Cases = 6.0Item-total Statistics Scale Scale Corrected Mean Variance Item- Squared Alpha if Item if Item Total Multiple if Item Deleted Deleted Correlation Correlation DeletedITEM_1 13.0000 6.4000 .5251 .64
14、71 .8472ITEM_2 13.1667 5.3667 .6757 .7500 .8116ITEM_3 12.3333 5.4667 .8333 .8588 .7642ITEM_4 13.5000 6.7000 .7481 .7857 .8093ITEM_5 12.6667 5.8667 .5922 .7143 .8333Reliability Coefficients 5 itemsAlpha = .8457 Standardized item alpha = .8609纸笔计算结果Question12345Total ScoreJoe3443519Sam4343317Sue233231
15、3Peg4453420Gil3243315Dot3232313i2 =.4722.6667.4722.2222.5833=16.1667, 2=7.4722 =2.7335 k=5 N=6【伍】 KR20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 转换数据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
16、 * 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_11.0000ITEM_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 * * * Determ
17、inant 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 = 5.0Item-total Statistics Scale Scale Corrected Mean Variance Item- Squared Alpha if Item if Item Total Multiple if Item Deleted Deleted Correlation Correlation
18、 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 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Question123456Total ScoreAYYYYNY5BYNNYNY3CYNNYNN2DNYNNNN1ENYNNYY3Pi.6.6.2.6.2.6qi.4.4.8.4.8.4(p)(q).24.24.16.24.16.24=2.8, 2=1.76 =1.3266 k=6 N=5