I. FillingBlanks（28%）

Fill in the blanks in the following table. Keepall results to 4 decimal places.  

Source	SS	df	MS	Number of obs	=	9
				F(1, 49302)	=	10
Model	1	2	3	Prob >  F	=	0.000
Residual	4	5	6	R-squared	=	0.0009
Total	10727.3883	7	8	Adj R-squared	=	11
				Root MSE	=	12

Y	Coef.	Std. Err.	t	P>|t|	[95% Conf.  Interval]
X	-.0017457	.0002662	13	0.000	15	17
_cons	5.689695	.002235	14	0.000	16	18

考点：回归分析结果阅读（28分）

1. 回归结果：离差					2. 模型显著性
Source	SS  平方和	df 自由度	MS 均方		Number of obs	=	49304
					F(1, 49302)	=	44.4118
Model	9.6546	1	9.6546		Prob > F	=	0
Residual	10717.7337	49302	0.2174		R-squared	=	0.0009
Total	10727.3883	49303	0.2176		Adj R-squared	=	0.0009
					Root MSE	=	0.4663
3.  关键回归结果：标准误、t检验值、t检验的p值，置信区间
Y	Coef.	Std.Err.	t		P>|t|	[95% Conf.Interval]
Xj	-0.0017457	0.0002662	-6.5579		0.000	-0.0023	-0.0012
_cons	5.689695	0.002235	2545.7248		0.000	5.6853	5.6941

II. Questions and Brief Answers （42%）  

1. Given a standardized normaldistribution (with a mean of 0 and a standard deviationof 1), what is the probabilitythat (8%)  

a. Z is less than 1.08?  

b. Z is greater than -0.21?  

c. Z is less than -0.21 or greater than the mean?  d. Z is less than-0.21 or greater than 1.08?  

考点：概率的概念，查累计标准分布表（Z值表）（8分）

SOLUTION:

a.     P(Z<1.08) =0.8599=85.99%

b.     P(Z>-0.21)=1-P(Z<=-0.21)=1-0.4207=0.5793

c.     P(Z<-0.21 orZ>0)=P(Z<-0.21)+P(Z>0)=P(Z<-0.21)+[1-P(Z<=0)]=0.4207+(1-0.5)=0.9207

d.     P(Z<-0.21 or Z>1.08)=P(Z<-0.21)+[1-P(Z<=1.08)]=0.4207+(1-0.8599)=0.5608

1  

2. Many manufacturing problems involve thematching of machine parts, such as shaftsthat fit into a valve hole. A particular design requires a shaft with a diameter of 22.000 mm, but shafts withdiameters between 21.990 mm and22.010 mm are  
acceptable. Suppose that the manufacturing processyields shafts with diameters normally distributed, with a mean of 22.002 mm and a standard deviation of 0.005mm. For this process, what is (8%)  

a. the proportion ofshafts with a diameter between 21.99 mm and 22.00mm?   

b. the probability that ashaft is acceptable?  

c. the diameter that will be exceeded by only2% of the shafts?  

d. What would be your answersin (a) through (c) if the standard deviation of the shaft diameters were 0.004 mm?  

考点：简单Z值计算，抽样与抽样分布；一般模型向正态分布模型的转化，以及结果的还原或呈现。（8分）

读题：

1.    设计要求为22.000mm，可接受的均值μ区间为[21.990, 22.010]；

2.    题中的样本是正态分布的，样本均值为22.002mm，标准差为0.005mm。

SOLUTION：

a.     转化为Z值，求概率。
Using equation: 普通的正态分布——>标准正态分布的一个变换

Z-values fordiameter=21.99mm and diameter=22.00mm are respectively -2.4 and -0.4.

P(21.99mm<=diameter<=22.00mm) = P(-2.4<=Z<=-0.4)=P (-0.4)-P (-2.4) = 0.3446- 0.0207=0.3239

So, the proportionof shafts with a diameter between 21.99 mm and 22.00 mm is 32.39%.

b.     A shaft is acceptable when its diameter is between 21.990mm and22.010mm.

Similar to thesolution in (a.),

P(21.990mm<=diameter<=22.010mm) =P(-2.4<=Z<=1.6) =P (1.6)-P (-2.4) =0.9515-0.0207=0.9308

So, the proportionthat a shaft is acceptable is 93.08%.

c.     For the diameter thatwill be exceeded by only 2% of the shafts, the Z-value represents the probabilityof 98%, which equals to 2.05(or 2.06, 2.05 is preferred).

Using equation:

So, the diameter that willbe exceeded by only 2% of the shafts is 22.01225mm.

d.     If the standarddeviation of the shaft diameters were 0.004 mm:

a)     For a., Z-values for diameter=21.99mmand diameter=22.00mm are respectively -3 and -0.32.

P(21.99mm<=diameter<=22.00mm)= P(-3<=Z<=-0.32) =P (-0.32)-P (-3) = 0.3745 - 0.00135 = 0.37315

So, the proportionof shafts with a diameter between 21.99 mm and 22.00 mm is 37.32%.

b)     For c., using equation:

So, thediameter that will be exceeded by only 2% of the shafts is 22.0102mm.

2