内容发布更新时间 : 2024/12/25 23:31:19星期一 下面是文章的全部内容请认真阅读。
Test Statistics性别(1=雄性;2=雌性)Chi-SquareadfAsymp. Sig.5.8811.015a. 0 cells (.0%) have expected frequencies less than5. The minimum expected cell frequency is 71.5. (4)作出推断:由上表可知 P=0.015 <α=0.05,故否定H0,接受HA认为野兔性别比例不符
合1:1的比例。
习题5.4 解:(1)H0:大麦F2代芒性状表型的其比率符合9:3:4的理论比率;
HA:其比率不符合9:3:4的理论比率; (2)选择显著水平为0.05
(3)经SPSS卡方分析得到如下结果:
芒性状表型(1=钩芒;2=长芒;3=短芒)Observed NExpected N123Total348115157620348.7116.4154.8Residual-.7-1.42.2
Test Statistics芒性状表型(1=钩芒;2=长芒;3=短芒)Chi-SquareadfAsymp. Sig..0492.976a. 0 cells (.0%) have expected frequencies less than5. The minimum expected cell frequency is 116.4. (4)作出推断:由上表可知 P=0.976 >α=0.05,故接受H0,否定HA,认为大麦F2代芒性状表
型比率符合9:3:4的理论比率。
16 / 43
习题5.5 解:(1)H0:这群儿童性别比合理;
HA:这群儿童性别比不合理; (2)选择显著水平为0.05
(3)经SPSS卡方分析得到如下结果:
性别(1=男孩;2=女孩)Observed NExpected N12Total421326747373.5373.5Residual47.5-47.5 Test Statistics性别(1=男孩;2=女孩)Chi-SquareadfAsymp. Sig.12.0821.001a. 0 cells (.0%) have expected frequencies less than5. The minimum expected cell frequency is 373.5. (4)作出推断:由上表可知 P=0.001<α=0.05,故否定H0,接受HA,认为这群儿童性别比不
合理。
习题5.6
解:(1)H0:两种苹果的耐储性差异不显著;
HA:两种苹果的耐储性差异显著; (2)选择显著水平为0.05;
(3)经SPSS卡方分析得到如下结果:
品种(1=国光;2=品种) * 腐烂情况(1=腐烂;2=未腐烂) CrosstabulationCount腐烂情况(1=腐烂;2=未腐烂)1品种(1=国1光;2=品种)2Total1416302186162348Total200178378
17 / 43
Chi-Square TestsValuePearson Chi-SquareaContinuity Correctiondf111Asymp. Sig.(2-sided).475.601.476Exact Sig.(2-sided)Exact Sig.(1-sided).510b.274.509Likelihood RatioFisher's Exact TestLinear-by-LinearAssociationN of Valid Cases.568.5083781.476.300a. Computed only for a 2x2 tableb. 0 cells (.0%) have expected count less than 5. The minimum expected count is14.13.(4)作出推断:由上表可知 P=0.274>α=0.05,所以接受H0 ,否定HA,即两种苹果的耐储性差异未达显著。
习题5.7
解: (1)H0:不同小麦品种与赤霉病的发生无显著关系;
HA:不同小麦品种与赤霉病的发生有显著关系; (2)选择显著水平为0.05;
(3)经SPSS卡方分析得到如下结果
植株患病情况(1=健株;2=病株) * 小麦品种(1=A;2=B;3=C;4=D;5=E) CrosstabulationCount小麦品种(1=A;2=B;3=C;4=D;5=E)1植株患病情况(1=健株;2=病株)Total12442785202460394993478355134376298674549450544Total22505002750 Chi-Square TestsValuePearson Chi-SquareContinuity CorrectionLikelihood RatioLinear-by-LinearAssociationN of Valid Cases379.68832.775275041.000.000420.671adf4Asymp. Sig.(2-sided).000 (4)作出推断:由上表可知 P值为0.000<α=0.05,故否定H0,接受HA,说明不同小麦品
种与赤霉病的发生有极显著的关系。
18 / 43
a. 0 cells (.0%) have expected count less than 5. Theminimum expected count is 90.73.习题5.8
解: (1)H0:灌溉方式与叶片衰老无关;
HA:灌溉方式与叶片衰老有关; (2)选择显著水平为0.05;
(3)经SPSS卡方分析得到如下结果
灌溉方式(1=深水;2=浅水;3=湿润) * 稻叶情况(1=绿叶;2=黄叶;3=枯叶) CrosstabulationCount稻叶情况(1=绿叶;2=黄叶;3=枯叶)1灌溉方式(1=深1水;2=浅水;3=2湿润)3Total146183152481279143037131636Total160205182547
Chi-Square TestsValuePearson Chi-SquareContinuity CorrectionLikelihood RatioLinear-by-LinearAssociationN of Valid Cases5.5354.51054741.237.0345.622adf4Asymp. Sig.(2-sided).229 (4)作出推断:由上表可知 P=0.229>α=0.05,故接受H0,否定HA说明水稻灌溉方式与叶片
衰老无关。
a. 0 cells (.0%) have expected count less than 5. Theminimum expected count is 8.78.
19 / 43