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(IELM313)sol_ielm313_hw3.pdf
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Solution to IELM313 Homework #3
Solution to Question 1.
Let X =(X1, ,X10) denote these 10 independent observations. Let R = 10 denote the sample size. Then by a point estimate of is
1 R
.= . Xi =0.792.
R
i=1
Its 95% con.dence interval is
.SS ... t1./2,R.1 ,.+ t1./2,R.1 ,
RR
1
where =1 . 95% = 0.05 and S2 = .R (Xi . .)2 .
R.1 i=1
By simple calculation we have S =0.052 and by checking the table of t-distribution we know that t1./2,R.1 =2.2622. Hence the 95% con.dence interval of is (0.755, 0.829). Point estimate of p is
1
p.= #{Xi 0.8} =0.5,
R
and its 95% con.dence interval is ..p.(1 . p.) .p.(1 . p.) .
p.. t1./2,R.1 ,p.+ t1./2,R.1 = (0.123, 0.877).
R . 1 R . 1
In order to set the number of replications, we pick R large enough such that t1./2,R.1 = z1./2 =1.96. Then halfwidth of the 95% con.dence interval based on n observations is S
.
z1./2
n In order to guarantee that the halfwidth is no more than 0.005, then we have
.2. S
n z1./2 416.
0.005 Hence we need 416 . 10 = 406 additional observations.
.p.(1.p.)
Similarly, halfwidth of 95% con.dence of p is z1./2 . Then
n.1
n .z1./2 .2
p.(1 . p.)+1 = 9605.
0.01
Hence we need 9605 . 10 = 9595 additional observations. 1
Solution to Question 2.
We .rst sort the data from the smallest to the largest. Then point estimate of q is the 90-th smallest, i.e., q.=0.886.
Moreover,
.0.9(1 . 0.9)
pl =0.9 . z1.0.05/2 =0.84,
99 and
.0.9(1 . 0.9)
pu =0.9+ z1.0.05/2 =0.96.
99 Then lower and upper limits of 95% con.dence interval of q are the 84-th and 96-th smallest, respectively, i.e., 95% con.dence interval is (0.880, 0.903).
Remark 1. Since the data in Question 2 is in an Excel spreadsheet, which is generated according to certain random distribution, its possible that the data .le will appear to be di.erent when you reopen it or even open it in di.erent computers. This is due to the algorithm and random generator used by Excel. Hence you .nal results of Question 2 could be slightly di.erent. However, the calculation of pl and pu is standard.
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