运用SPSS进行信度分析报告
发布时间:2019-10-22 10:06:42
发布时间:2019-10-22 10:06:42
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)
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:
Person | A | B | C | D | E | F | G | H | I | J | σ | |
Oct | 18 | 16 | 5 | 13 | 15 | 16 | 12 | 5 | 8 | 10 | 11.8 | 4.4226 |
Apr | 18 | 18 | 6 | 16 | 17 | 16 | 14 | 5 | 7 | 11 | 12.8 | 4.8744 |
X1 X2 | 324 | 288 | 30 | 208 | 255 | 256 | 168 | 25 | 56 | 110 | ||
N=10
复本信度(Alternate-form Reliability)
Correlations
纸笔计算结果:
Person | A | B | C | D | E | F | G | H | I | J | σ | |
Form A | 16 | 12 | 14 | 10 | 9 | 11 | 13 | 9 | 16 | 12 | 12.2 | 2.4413 |
Form B | 15 | 12 | 15 | 10 | 10 | 12 | 14 | 9 | 16 | 13 | 12.6 | 2.289 |
XA XB | 240 | 144 | 210 | 100 | 90 | 132 | 182 | 81 | 256 | 156 | ΣXAXB=1591 | |
N=10
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Bivariate
Correlations
纸笔计算结果
Ⅰ. 计算两个”半测验”的相关
Student | Joe | Sam | Sue | Peg | Gil | σ | |
Odds | 2 | 1 | 2 | 1 | 1 | 1.4 | .4899 |
Evens | 3 | 2 | 3 | 1 | 2 | 2.2 | .7483 |
X1X2 | 6 | 2 | 6 | 1 | 2 | X1X2 =17 | |
N=5
Ⅱ 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 Coefficients
N of Cases = 5.0 N of Items = 6
Correlation 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
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
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
Correlation 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】
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 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
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
Question | 1 | 2 | 3 | 4 | 5 | Total Score |
Joe | 3 | 4 | 4 | 3 | 5 | 19 |
Sam | 4 | 3 | 4 | 3 | 3 | 17 |
Sue | 2 | 3 | 3 | 2 | 3 | 13 |
Peg | 4 | 4 | 5 | 3 | 4 | 20 |
Gil | 3 | 2 | 4 | 3 | 3 | 15 |
Dot | 3 | 2 | 3 | 2 | 3 | 13 |
σi2 = | .4722 | .6667 | .4722 | .2222 | .5833 | |
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
ITEM_1 | ITEM_2 | ITEM_3 | ITEM_4 | ITEM_5 | ITEM_6 | |
ITEM_1 | 1.0000 | |||||
ITEM_2 | -.6667 | 1.0000 | ||||
ITEM_3 | .4082 | .4082 | 1.0000 | |||
ITEM_4 | 1.0000 | -.6667 | .4082 | 1.0000 | ||
ITEM_5 | -.6124 | .4082 | -.2500 | -.6124 | 1.0000 | |
ITEM_6 | .1667 | .1667 | .4082 | .1667 | .4082 | 1.0000 |
* * * Warning * * * Determinant of matrix is close to zero: 8.426E-36
Statistics based on inverse matrix for scale ALPHA
are meaningless and printed as .
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
Question | 1 | 2 | 3 | 4 | 5 | 6 | Total Score |
A | Y | Y | Y | Y | N | Y | 5 |
B | Y | N | N | Y | N | Y | 3 |
C | Y | N | N | Y | N | N | 2 |
D | N | Y | N | N | N | N | 1 |
E | N | Y | N | N | Y | Y | 3 |
Pi | .6 | .6 | .2 | .6 | .2 | .6 | |
qi | .4 | .4 | .8 | .4 | .8 | .4 | |
(p)(q) | .24 | .24 | .16 | .24 | .16 | .24 | |