豆子編製 3/10/20181心理測驗心理測驗 SPSS 上機：信度分析上機：信度分析資料輸入Data 輸入頁變項定義頁豆子編製 3/10/20182豆子編製 3/10/20183豆子編製 3/10/20184Reliability Analysis本次課題包含：如何進行各種信度分析1. 再測信度（Test-Retest Reliability）2. 複本信度（Alternate-form Reliability）3. 折半信度（Split-half Reliability）4. 內部一致性（Internal Consistency Coefficient） 【計算 係數】5. KR 20【壹】再測信度（Test-Retest Reliability）步驟一 按【Analyze】【Correlate】【Bivariate】某空間性向測驗有 20 題單選題，分別在十月與第二年四月施測同一組 10 名學生，以下是測驗結果，請計算信度。PersonABCDEFGHIJOct18165131516125810Apr18186161716145711豆子編製 3/10/20185步驟二 會出現下面的對話框，將左邊兩變項選入右邊Variables內，在Correlation Coefficients方盒內選取Pearson ；在Test of Significance方盒內選取Two-tailed ；勾選最下面的Flag significant correlations ，之後按鍵。補充 若想呈現平均及標準差則可在按鍵前按進入下個對話框，在 Statistics 的方盒內選取Means and standard deviations ，按繼續。Correlations豆子編製 3/10/20186Descriptive Statistics11.80004.6619010 12.80005.1380910OCT APRMeanStd. DeviationNCorrelations1.972* .000 1010 .972*1 .000. 1010Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) NOCTAPROCTAPRCorrelation is significant at the 0.01 level (2-tailed).*. 紙筆計算結果：PersonABCDEFGHIJXOct1816513151612581011.8 4.4226Apr1818616171614571112.8 4.8744X1 X2324288302082552561682556110X1X2 =1720N=1097.97228.557521.2104.151172 )8744. 4)(4226. 4()8 .12)(8 .11()10/1720(XXr【貳】複本信度（Alternate-form Reliability）做法與再測信度相同，請參照。Correlations某自我概念量表，有 20 題是非題，題本 A 與 B 分別施測同一組 10 個人， 分數愈高表示愈具有正向自我概念，以下是施測結果，請計算信度。PersonABCDEFGHIJForm A161214109111391612Form B1512151010121491613豆子編製 3/10/20187Descriptive Statistics12.20002.5733710 12.60002.4129310FORM_A FORM_BMeanStd. DeviationNCorrelations1.963*.000 1010.963*1.000. 1010Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) NFORM_AFORM_BFORM_AFORM_BCorrelation is significant at the 0.01 level (2-tailed).*. 紙筆計算結果：PersonABCDEFGHIJXForm A16121410911139161212.22.4413Form B151215101012149161312.62.289XA XB2401442101009013218281256156XAXB=1591N=1096.9627.5883798. 572.1531 .159 )2891. 2)(4413. 2()6 .12)(2 .12()10/1591(xxr【參】折半信度（Split-half Reliability）6 題的隨堂測驗施測 5 位學生， Y 表示答對，N 表示答錯，以 下是測驗結果，請計算信度。PersonItem 1Item 2Item 3Item 4Item 5Item 6JoeYYYYNYSamYNNYNYSueYYNYYYPegNYNNYNGilNYNNYY豆子編製 3/10/20188步驟一 輸入資料步驟二 轉換資料為數字按【Transform】【Recode】【Into Same Variables】豆子編製 3/10/20189出現下面的對話框後將左邊方格內 item1item6 選至右邊 String Variables 內後點選鍵出現下列對話框後，將”N”定義為”0”，將”Y”定義為”1”後按鍵豆子編製 3/10/201810之後便會將資料轉換成下面的數字步驟三 將string的屬性改為numeric豆子編製 3/10/201811步驟四 計算奇數題與偶數題的和按【Transform】【Compute】即出現下面的對話框豆子編製 3/10/201812（此為奇數題的範例，偶數題亦然）結束後便會在 spss Data Editor 對話框中出現奇數題和偶數題的和豆子編製 3/10/201813步驟四 執行 BivariateCorrelationsDescriptive Statistics1.4000.547725 2.2000.836665ODDS EVENSMeanStd. DeviationN豆子編製 3/10/201814Correlations1.873 .053 55 .8731 .053. 55Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) NODDSEVENSODDSEVENS紙筆計算結果. 計算兩個”半測驗”的相關 StudentJoeSamSuePegGilXOdds212111.4.4899Evens323122.2.7483X1X262612X1X2 =17N=587.8729.3665922.08. 34 . 3)7483)(.4899(.)2 . 2)(4 . 1 ()5/17(XXr 校正相關係數為折半信度Spearmen-Brown prophesy formula 史比校正公式 (當兩個半測驗變異數相等時使用)(1(1)(XXXX XXcurrentNcurrentNNEW93.9321.8729.17458. 1)8729)(.12(1)8729. 0)(2(XXNEWGuttman prophesy formula 哥德曼校正公式 (當兩個半測驗變異數不等時使用)1 (2222XEO OEr888. 0)8 . 1 7 . 03 . 01 (2)3416. 1 ()8367. 0()5477. 0(1 (2222 OENEW*折半信度* 折半信度也可直接使用 SPSS 計算 豆子編製 3/10/201815步驟一 輸入資料步驟二 按【Analyze】【Scale】【Reliability Analysis】將左邊方格內的變項依所需次序分前後半選入右邊 items 的方格內，在左下角的 Model 框中選取 Split-half 後按鍵，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 = .8889 Unequal-length Spearman-Brown = .93213 Items in part 1 3 Items in part 2Alpha for part 1 = -2.5000 Alpha for part 2 = .0000標準的折半信度算法：先求各題的難度，再依難度加以排序，再施行上述的方 法豆子編製 3/10/201816Descriptive Statistics53.60 51.20 53.60 54.80 53.60 54.805ITEM_1 ITEM_3 ITEM_5 ITEM_2 ITEM_4 ITEM_6 Valid N (listwise)NSumMean透過平均值可看出其難度平均高的難度低 平均低的難度高排序由難度低到高 2 6 1 4 5 3，在丟入變項時依單偶分為： 2 1 5、6 4 3 兩組，排列數據時 前後排列。 CorrelationsCorrelations1.000.1.000 55.00011.000. 55Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) NSFSFCorrelations* 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 = .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 range