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應數系高等統計學習題四
1. Let two independent random samples, each of size ten, from two independent
normal distributions. N ( 1 ,  2 ) and N ( 2 ,  2 ) yield x  4.8, s12  8.64,
y  5.6, s22  7.88 .
(1) Please test the hypothesis H 0 : 1  2 vs. H1:1  2 .
(2) Find a 95 percent confidence interval for 1  2 .
2. The following random samples are measurements of the heat-producing capacity
(in millions of calories per ton)of specimens of coal from two mines:
Mine 1:8,380 8,180 8,500 7,840 7,990
Mine 2:7,660 7,510 7,910 8,070 7,790
(1) Use the 0.05 level of significance to test whether the difference between the
means of these two samples is significant.
(2) Find a 95 percent confidence interval for the difference of two means.
3. Interval Estimation and Testing Hypothesis Students may choose taking statistics
courses from professor A or B. The final written examination is the same for class
A and class B. If 13 students in class A and 11 students in class B. Let  A and
B denote the population mean of class A and class B, respectively, and  A2 and
 B2 denote the variance of class A and class B, respectively. The sample average
and standard deviation of each class are Class A: xA  84, s A  4 ,
Class B: xB  77, sB  6 .(Use   0.1 )
(1) Construct an appropriate statistical hypothesis of testing to judge whether
 A  B .
(2) Find a 95 percent confidence interval for the difference of two means.
4. China Airline researcher wants to understand that this company's jade emperor
system's result. The researcher extracts n=100 day-long localization record, which
is the random sample, and estimates that the mean value which how many people
subscribe in advance sit actually but have not ridden by Taipei to Kaohsiung flight
every day 4:10PM. Obtains the following data by the record
Booking actually but
have not ridden ( X )
0
1
2
3
4
5
6
Number of days
20
37
23
15
4
0
1
Use the Chi-square test to test whether X is Poission distribution.
(Use   0.1 )
5. Some researcher is studying the average life of one kind of the tire. Must first
confirm this tire's attrition distance in kilometer in the research first stage whether
to become the normal distribution. Its experiment's result is as follows
Attrition distance in
kilometer(interval)
Intermediate points
Numbers
 23,000
22,500
33
23,000~24,000
23,500
31
24,000~25,000
24,500
66
25,000~26,000
25,500
100
26,000~27,000
26,500
84
27,000~28,000
27,500
57
 28,000
28,500
29
Does this one kind of the tire attrition number become the normal distribution?
2
2
(6)  16.81,  0.01
(4)  13.24 )
(Use   0.01 ,  0.01