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1、单选题1 .单样本等级资料比较时应选用()A.单样本t检验B.单样本检验C.单样本2检验D.符号秩和检验2 .两组或多组有序分类变量资料的比较宜用()A. t检验B.直线回归分析C.秩和检验或Ridit分析D.四格表2检验3 .在进行成组设计两样本秩和检验时,以下哪种检验假设是正确的()A.HO:两总体均数相等 B.HO:两样本均数相等 C.HO:两总体分布相同 D.HO:两样本分布相同H1:两总体均数不等H1:两样本均数不等H1:两总体分布不同H1:两样本分布不同4 . Ridit分析适用于(A.偏态分布的数值变量资料的假设检验B.方差不齐的几种正态分布资料的比较C有序分类变量资料的比较D.
2、多个样本率的比较5 . Ridit分析属于()A.参数检验B.非参数检验C.秩和检验D.相关回归分析6 .欲比较三种药物的疗效(无效、好转、显效、痊愈)孰优孰劣,最好选择()A.r检验B.方差分析C*检验(两样本配对的计数资料卡方检验)D.秩和检验(多样本的计数资料的秩和检验)7 .等级资料的比较宜用()A.t检验8 .秩和检验C.F检验D.四格表/检验8.在两独立样本比较的秩和检验中,实验组的观察值为0, 0, 7, 14, 32, 40,对照组的观察值为0, 1, 2, 4, 4,8.编秩中零值的秩应分别编为().A.2, 3, 1B.1.5, 1.5, 3C.2, 2, 2D. 2, 3
3、, 4判断题1 .单样本等级资料的游程检验可以检验单样本等级资料的取值是否是随机的。()2 .当两个样本的等级资料进行比较,且各组总频数均N20时,使用Ridit分析。()Chi-Square TestsValuedfAsymp. Sig. (2-sided)Exact Sig. (2- sided)Exact Sig. (1- sided)Point ProbabilityPearson Chi-Square3.895a4.420,431Likelihood Ratio3.7534.440,460Fishers Exact Test3.690.459Linear-by-Linear.248,
4、275.143.035Association1.332b1McNemar-BowkerTest1.7593,624N ofValid Cases78a. 0 cells (0.0%) have expected countless than 5. The minimum expected count is 5.90. b.The standardized statistic is 1.154.Symmetric MeasuresValueAsymp. Std. Error3Approx. TbApprox. Sig.Exact Sig.Measure of Agreement Kappa N
5、ofValid Cases.15478.0831.937.053.070a. Not assuming the null hypothesis.b. Using the asymptotic standard error assuming the null hypothesis.下列说法正确的是()A.两种抗体检测方法检测结果无关联B.乳胶凝集法的检出率高于免疫荧光法C.免疫荧光法的检出率等于乳胶凝集法D. Kappa 值等于0.154E.两种检测有一致性3 .两组独立样本的等级资料进行比较时,若样本容量较小选用秩和检验。()4 .多样本有序分类资料的Kruskal Wallis H秩和检验一
6、般是在各组样本量很大的时候使用,且结论可以得出多组间总 的差别和两两间是否有差别。()5 .多样本有序分类资料的Kruskal Wallis H秩和检验一般是在各组样本量较小的时候使用,且结论只得出多组间总的 差别,若要知道两两间是否有差别需要进一步检验。() 简答题1 .两组或多组等级资料的比较,为什么不能用/检验,而用秩转换的非参数检验?2 .样本(对比组)与总体(标准组)比较的Ridit分析,可以采用什么方法?3 . Bowker检验的基本思想是什么?参考答案Bowker检验的基本思想是检验平方表主对角线上下对称格子中的频数是否相等。若上下对称格子中的频数 相等,则认为两种方法或两种检测
7、手段的效果没有差别。若上下对称格子中的频数不相等,经检验,计算的统计量大 于或等于统计量的临界值时,则认为两种检测有差别。4 .游程检验的基本思想是什么?参考答案游程检验可以检验单样本等级资料的取值是否是随机的。依时间或其他顺序排列的有序数列中,具有相同 属性的事件或符号的连续部分称为一个游程,每个游程含有事件或符号的个数称为游程的长度。在一个有序数列中, 游程的个数记为r,游程的长度记为L。单选题1 .三种类型的产妇产后泌乳量的人数分布数据见表1,统计分析结果如下:表1三种产妇在产后一个月内的泌乳量无少多早产303610足月产13229214过期产101434Descript rvesSta
8、tisticStd. Error早产俎Mean.397514.027497095% Confidence Interval for MeanLower Bound.342933Upper Bound,452096Descript vvesStatisticStd Error足月产蛆Mean95% Confidence Interval for MeanLower BoundUpper Bound.508879.491158.526599.0090283Descript ivesStatisticStd. Error过期产组Mean,543119.035689295% Confidence In
9、terval for MeanLower Bound.471652Upper Bound,614585上述资料数据类型是()A.计量资料B.计数资料C.等级资料D.数值型变量资料单选题2 .三种类型的产妇产后泌乳量的人数分布数据见表1,统计分析结果如下: 表1三种产妇在产后一个月内的泌乳量无少多早产303610足月产13229214过期产101434Descript rvesStatisticStd. Error早产俎Mean.397514.027497095% Confidence Interval for MeanLower Bound.342933Upper Bound,452096De
10、scripth/esStatisticStd Error足月产俎 Mean95% Confidence IntervalLower Boundfor Mean.一Upper Bound.508879.491158.526599.0090283Descript ivesStatisticStd. Error过期产组Mean.543119.035689295% Confidence Interval for MeanLower Bound,471652Upper Bound,614585上述资料采用的分析方法是()A. Kruskal-Wallis H 检验B. Ridit 分析C.秩相关分析D.
11、 Mann-Whitney U 检验多选题3 .三种类型的产妇产后泌乳量的人数分布数据见表1,统计分析结果如下: 表1三种产妇在产后一个月内的泌乳量无少多早产303610足月产13229214过期产101434Descript rvesStatisticStd. Error早产俎Mean.397514.027497095% Confidence Interval for MeanLower Bound.342933Upper Bound,452096Descripth/esStatisticStd Error足月产俎 Mean95% Confidence IntervalLower Bound
12、for Mean.一Upper Bound.508879.491158.526599.0090283Descript ivesStatisticStd. Error过期产组Mean.543119.035689295% Confidence Interval for MeanLower Bound,471652Upper Bound,614585下列说法正确的是()A.三种产妇产后一个月内的泌乳量没有差异B.过期产和足月产产妇在产后一个月内的泌乳量没有差异C.过期产产妇产后一个月内的泌乳量高于早产产妇D.足月产产妇产后一个月内的泌乳量高于早产产妇E.足月产产妇产后一个月内的泌乳量高于过期产产妇判
13、断题4 .三种类型的产妇产后泌乳量的人数分布数据见表1,统计分析结果如下: 表1三种产妇在产后一个月内的泌乳量无少多早产303610足月产13229214过期产101434Descript rvesStatisticStd. Error早产俎Mean.397514.027497095% Confidence Interval for MeanLower Bound.342933Upper Bound,452096Descripth/esStatisticStd Error足月产组Mean95% Confidence Interval for MeanLower BoundUpper Bound
14、.508879.491158.526599.0090283Descript ivesStatisticStd. Error过期产组Mean.543119.035689295% Confidence Interval for MeanLower Bound,471652Upper Bound,614585上述材料中所使用的统计分析方法适用于等级资料的比较且样本量较大,其结论可以得出多组间总的差别和两两间是 否有差别。()简答题5 .言种类型的产妇产后泌乳量的人数分布数据见表1,统计分析结果如下:表1三种产妇在产后一个月内的泌乳量无少多早产303610足月产13229214过期产101434Des
15、cript rvesStatisticStd. Error早产俎Mean.397514.027497095% Confidence Interval for MeanLower Bound.342933Upper Bound,452096Descripth/esStatisticStd Error足月产组Mean95% Confidence Interval for MeanLower BoundUpper Bound.508879.491158.526599.0090283Descript ivesStatisticStd. Error过期产组Mean.543119.035689295% C
16、onfidence Interval for MeanLower Bound,471652Upper Bound,614585上述材料中所使用的统计分析方法的基本思想是什么?单选题1.某实验室分别用乳胶凝集法和免疫荧光法对78名可疑系统红斑狼疮患者血清中抗核抗体进行测定,数据见表1, 分析如下:表1 78名可疑系统红斑狼疮患者血清中抗核抗体测定结果 乳胶凝集法免疫灾犬法+1 4+ + +1486+ +711?1089Chi-Square TestsValuedfAsymp. Sig. (2-sided)Exact Sig. (2- sided)Exact Sig. (1- sided)Poin
17、t ProbabilityPearson Chi-Square3.895a4.420.431Likelihood Ratio3.7534.440.460Fishers Exact Test3.690.459Linear-by-LinearAssociation1.332b1.248.275.143,035McNemar-BowkerTest1.7593.624N ofValid Cases78a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 5.90.b.The standardiz
18、ed statistic is 1.154.Symmetric MeasuresValueAsymp. Std. Error3Approx. TbApprox. Sig.Exact Sig.Measure of AgreementKappaN ofValid Cases.15478.0831.937.053.070a. Not assuming the null hypothesis.b. Using the asymptotic standard error assuming the null hypothesis.上述资料数据类型是()A.计量资料B.计数资料C.等级资料D.数值型变量资料
19、6.第13章计数资料RxC表Pearson卡方检验可以用来比较多组构成比或率有无差别,此处KxK方表能否说明两种方 法各个等级构成比或率有无差别?单选题2 .某实验室分别用乳胶凝集法和免疫荧光法对78名可疑系统红斑狼疮患者血清中抗核抗体进行测定,数据见表1,分析如下:表1 78名可疑系统红斑狼疮患者血清中抗核抗体测定结果免投英光法乳胶凝集法+ +1486+ +711+1089Chi-Square TestsValuedfAsymp. Sig. (2-sided)Exact Sig. (2- sided)Exact Sig. (1- sided)Point ProbabilityPearson
20、Chi-Square3.895a4.420.431Likelihood Ratio3.7534.440.460Fishers Exact Test3.690.459Linear-by-LinearAssociation1.332b1.248.275.143.035McNemar-BowkerTest1.7593.624N ofValid Cases78a. 0 cells (0.0%) have expected countless than 5. The minimum expected count is 5.90. b.The standardized statistic is 1.154
21、.Symmetric MeasuresValueAsymp. Std. Error3Approx. TbApprox. Sig.Exact Sig.Measure of AgreementKappaN ofValid Cases.15478.0831.937.053.070a. Not assuming the null hypothesis.b. Using the asymptotic standard error assuming the null hypothesis.上述资料的设计类型是()A.成组设计B.完全随机设计C配对设计D.交叉设计单选题3 .某实验室分别用乳胶凝集法和免疫荧
22、光法对78名可疑系统红斑狼疮患者血清中抗核抗体进行测定,数据见表1, 分析如下:表1 78名可疑系统红斑狼疮患者血清中抗核抗体测定结果乳胶凝集法免疫灰光法+ + + +1486711+1089Chi-Square TestsValuedfAsymp. Sig. (2-sided)Exact Sig. (2- sided)Exact Sig. (1- sided)Point ProbabilityPearson Chi-Square3.895a4.420.431Likelihood Ratio3.7534.440.460Fishers Exact Test3.690.459Linear-by-L
23、inearAssociation1.332b1.248.275.143.035McNemar-BowkerTest1.7593.624N ofValid Cases78a. 0 cells (0.0%) have expected countless than 5. The minimum expected count is 5.90. b.The standardized statistic is 1.154.Symmetric MeasuresValueAsymp. Std. Error3Approx. TbApprox. Sig.Exact Sig.Measure of Agreemen
24、tKappaN ofValid Cases.15478.0831.937.053.070a. Not assuming the null hypothesis.b. Using the asymptotic standard error assuming the null hypothesis.欲比较甲乙两种方法抗体检测结果等级分布是否相同,选择的统计分析方法是()A.Pearson/检验B.Kappa 检验C.Fisher确切概率法D.McNemar-Bowker 检验单选题4.某实验室分别用乳胶凝集法和免疫荧光法对78名可疑系统红斑狼疮患者血清中抗核抗体进行测定,数据见表1, 分析如下:表
25、1 78名可疑系统红斑狼疮患者血清中抗核抗体测定结果乳胶凝集法免疫灵光法+1486+7115f1089Chi-Square TestsValuedfAsymp. Sig. (2-sided)Exact Sig. (2- sided)Exact Sig. (1- sided)Point ProbabilityPearson Chi-SquareLikelihood RatioFishers Exact TestLinear-by-LinearAssociationMcNemar-BowkerTestN ofValid Cases3.895a3.7533.6901.332b1.759784413.
26、420,440.248.624.431.460,459.275.143,035a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 5.90. b.The standardized statistic is 1.154.Symmetric MeasuresValueAsymp. Std. Error3Approx. TbApprox. Sig.Exact Sig.Measure of Agreement Kappa N ofValid Cases.15478.0831.937.053.0
27、70a. Not assuming the null hypothesis.b. Using the asymptotic standard error assuming the null hypothesis.欲比较甲乙两种方法抗体检测结果的一致性,选择的统计分析方法是()A.Pearson %2 检验B.Kappa 检验C.Fisher确切概率法D.McNemar 检验 单选题5.某实验室分别用乳胶凝集法和免疫荧光法对78名可疑系统红斑狼疮患者血清中抗核抗体进行测定,数据见表1, 分析如下:表1 78名可疑系统红斑狼疮患者血清中抗核抗体测定结果免投英光法乳胶凝集法+ +1486711+10
28、89Chi-Square TestsValuedfAsymp. Sig. (2-sided)Exact Sig. (2- sided)Exact Sig. (1- sided)Point ProbabilityPearson Chi-Square3.895a4.420.431Likelihood Ratio3.7534.440.460Fishers Exact Test3.690.459Linear-by-LinearAssociation1.332b1.248.275.143.035McNemar-BowkerTest1.7593.624N ofValid Cases78a. 0 cells
29、 (0.0%) have expected countless than 5. The minimum expected count is 5.90. b.The standardized statistic is 1.154.Symmetric MeasuresValueAsymp. Std. Error3Approx. TbApprox. Sig.Exact Sig.Measure of AgreementKappaN ofValid Cases.15478.0831.937.053.070a. Not assuming the null hypothesis.b. Using the asymptotic standard error assuming the null hypothesis.欲分析甲乙两种方法抗体检测结果是否存在关联,选择的统计分析方法是()A.Pearson 为2检验B.Kappa 检验C.Fisher确切概率法D.McNemar 检验多选题6.某实验室分别用乳胶凝集法和免疫荧光法对78名可疑系统红斑狼疮患者血清中抗核抗体进行测定,数据见表1, 分析如下:表1 78名可疑系统红斑狼疮患者血清中抗核抗体测定结果免疫荧光法乳胶凝集法+ 11486+ +711广. + .1089
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