某商场准备在商场内安装充电式应急照明灯,通过招标收到3家照明灯生产商的投标。该商场从3家生产商提供的应急照明灯样品中各随机抽取了5个进行检验,测得各个样品充电后可持续照明的时间长度(小时)数据如下:
若显著性水平α=0.05,检验3个生产商各自的应急照明灯的平均持续照明时间有无差异。您可能感兴趣的试卷
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假设检验中拒绝域为()
A.Z<-1.96
B.Z>1.96
C.Z>1.65
D.Z<-1.65
一家电器销售公司的管理人员认为,每月的销售额是广告费用的函数,并想通过广告费用对月销售额作出估计。用电视广告费用作自变量,销售额作因变量进行回归分析。下表给出了相关的分析结果,
下列说法正确的是()
A.广告费用每增加$1,000, 销售额增加$1.420 211
B.广告费用每增加$1,000, 销售额增加$1420. 211
C.广告费用每增加$1420.211, 销售额增加$1000
D.广告费用每增加$1, 销售额增加$363.689 1
A.组间平方和除以组内平方和
B.组内均方和除以组间均方
C.组间平方和除以总平方和
D.组间均方和除以组内均方
A.(1.645,1.96)
B.(0.40,0.50)
C.(0.45,0.55)
D.(0.419,0.481)
A.F0.05,10,20 = 1/F0.95,10,20
B.F0.05,10,20 = 1/F0.05,20,10
C.F0.95,10,20 = 1/F0.95,20,10
D.F0.95,10,20 = 1/F0.05,20,10
A.第一类错误被称为弃真错误
B.第一类错误被称为纳伪错误
C.第二类错误是原假设不正确,但却被接受的错误
D.第一类错误是原假设正确,但却被拒绝的错误
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The number of class intervals in a frequency distribution is usually between ().
某厂某年职工劳动生产率为20000元,是历史最高水平的1.2倍。这里的1.2倍是()
以2000年a0为最初水平,2016年an为最末水平,计算钢产量的年平均发展速度时,须开()
A politician who is running for the office of premier in a state with 3 million registered voters commissions a survey. In the survey, 53.6% of the 5000 registered voters interviewed say they plan to vote for him. The population of interest is the()
A cumulative frequency distribution would provide().
最能反映离散程度的指标是()
Simon Arnett, Director of Human Resources, is exploring the causes of employee absenteeism at Buderim Bottling during the last operating year (1 January 2005 to 31 December 2005). The average number of absences per employee, calculated from the personnel data of all employees, is a ().
One advantage of a stem and leaf plot over a frequency distribution is that ().
If the individual class frequency is divided by the total frequency, the result is the ().
什么是季节变动?研究它的意义何在?如何测定季节变动?