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雷达卡
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
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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.
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