MATH 448 confidence intervals discussion

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1. Let X1; X2; :::; Xn be iid with density

f (x; ) =  3x 23 0  x  0 otherwise where  > 0 is unknown.

a. Show that X(n) is a pivotal quantity.

b. Using (a), Önd the shortest and longest (1   ) 100% conÖdence intervals

for .

c. Using (a), Önd a (1   ) 100% lower conÖdence interval and a (1   ) 100%

upper conÖdence interval for . Compare with (b).

d. Refer to (b). For each of the Öve samples given below, Önd the shortest

and longest 40% conÖdence interv

MATH448 Spring 2019 Homework 2 Due: 02/20/2019 1. Let X1 ; X2 ; :::; Xn be iid with density 3x2 3 f (x; ) = 0 where > 0 is unknown. a. Show that 0 x otherwise X(n) is a pivotal quantity. b. Using (a), …nd the shortest and longest (1 ) 100% con…dence intervals for . c. Using (a), …nd a (1 ) 100% lower con…dence interval and a (1 ) 100% upper con…dence interval for . Compare with (b). d. Refer to (b). For each of the …ve samples given below, …nd the shortest and longest 40% con…dence intervals. Sample 1 0:923333203 0:978844113 0:599697742 Sample 2 0:739088058 0:849384877 0:624944651 Sample 3 0:547121685 0:572480989 0:940479844 Sample 4 0:585789623 0:90503358 0:807560687 Sample 5 0:770860022 0:868601028 0:740796493 e. How many of the …ve shortest con…dence intervals in (d) contained = 1? How many of the …ve longest con…dence intervals in (d) contained = 1? Compare these to their respective expected numbers. 2. Let X1 ; X2 ; :::; Xn be iid with density ( 2x x2 exp f (x; ) = 0 where > 0 is unknown. a. Show that x 0 x<0 d X12 = exp ( ) b. Show that n 2X 2 d Xi = 2 2n i=1 c. Using (b), construct a (1 ) 100% con…dence interval for . d. Refer to (c). Given X1 = 0:120458509; X2 = 0:726166173; X3 = 0:549261897, …nd a 95% con…dence interval for . 1 e. Using (b), construct a (1 ) 100% lower con…dence interval for . f. Using the result obtained in (e) and the data in (d), …nd a 98% lower con…dence interval for . 3. Let X1 ; X2 ; ::::; Xn be iid with density jxj x otherwise 2 f (x; ) = 0 a. Show that max X(1) ; X(n) is a pivotal quantity. b. Using (a), …nd the shortest and longest (1 ) 100% con…dence intervals for . c. Using (a), …nd a (1 ) 100% lower con…dence interval and a (1 ) 100% upper con…dence interval for . Compare with (b). d. For each of the samples listed below, …nd the shortest and longest 60% con…dence intervals. Sample 1 4:991281347 3:649769995 1:462805615 4:172191229 Sample 2 1:313565183 3:449088511 0:652047537 4:710386714 Sample 3 1:453151161 3:153812016 0:177819405 3:874609452 Sample 4 2:162890444 3:569092571 3:045221995 2:483571422 Sample 5 3:259516432 4:020841008 1:139512398 1:057256518 e. How many of the shortest con…dence intervals in (d) contain the value = 5? How many of the longest con…dence intervals in (d) contain the value = 5? Compare to their respective expected numbers. 4. Recall the de…nition of a T random variable with degrees of freedom. Z T =q W d d where Z = N (0; 1), W = 2 and Z and W are independent. In this exercise, you will derive the density of T in the following steps. a. Show that for w 0 and t 2 R lim "!0 P (t T t + "jW = w) 1 =p " 2 exp w wt2 2 This establishes that the conditional density of T given W = w is fT jW (tjw) = p 1 2 2 exp w wt2 2 for 1 < t < 1. b. Now recall that for all t 2 R Z1 Z1 fT (t) = fT;W (t; w) dw = fT jW (tjw) fW (w) dw 0 0 Use this fact, the result in (a) and the density of W to show that the density of T is +1 2 fT (t) = p for 1+ ( +1) 2 t2 2 1 < t < 1. d c. Refer to the result you found in (b). Suppose T = t . Show that if 0< 1 E (max (0; T )) = E (max (0; T )) = 1 and therefore E (T ) does not exist. d d. Refer to the result obtained in (b). Again suppose T = t . Show that if > 1, E (T ) = 0. e. Refer to the results in (c) and (d). Explain why the following does not necessarily hold. By independence of Z and W , p E (T ) = E (Z) E p =0 W Where do you think the problem with this is? f. Stirling’s Formula is the following. lim p t!1 (t) 2 tt 1 2 exp ( t) =1 Using Stirling’s Formula, show that the density you derived in (b) converges pointwise to the standard normal density function as ! 1. That is for each t 2 R, show that lim p !1 +1 2 1+ ( +1) 2 t2 2 3 1 = p exp 2 t2 2 5. Happy Gilmore’s medium cheesesteaks are becoming a major health concern for adults over 40 years old. The number of fat calories per cheesesteak is known to be normally distributed with unknown mean and unknown variance 2 . Seven cheesesteaks were randomly taken right out of Happy Gilmore’s for testing at the lab, and the number of fat calories were recorded for each. They were 516:243, 530:000, 537:219, 532:621, 520:094, 538:824 and 543:212. a. Construct a 95% con…dence interval for . b. Refer to (a). Would you say that the mean number of fat calories per cheesesteak is di¤erent than 530? c. Construct a 90% lower con…dence interval for . d. Refer to (c). Would you say that the mean number of fat calories per cheesesteak is greater than 500? 6. Happy Gilmore’s medium cheesesteaks are becoming a major health concern for adults over 40 years old. The number of fat calories per cheesesteak is known to be normally distributed with unknown mean and unknown variance 2 . Seven cheesesteaks were randomly taken right out of Happy Gilmore’s for testing at the lab, and the number of fat calories were recorded for each. They were 516:243, 530:000, 537:219, 532:621, 520:094, 538:824 and 543:212. a. Construct a 95% con…dence interval for b. Refer to (a). Would you say that found in (a)? 2 2 . 6= 225 based on the interval you c. Construct a 95% upper con…dence interval for . d. Refer to (c). Would you say that in (c)? < 17 based on the interval you found 7. A penny is tossed 148 times and shows heads 68 times. a. Construct an approximate 98% con…dence interval for the probability of the penny showing heads. b. Refer to (a). Do you think the coin is biased (i.e, the probability of showing heads is di¤erent than one half)? 8. A penny is tossed 148 times and shows heads 68 times. A dime is tossed 140 times and shows heads 72 times. Let p1 be the probability that the penny shows heads and p2 be the probability that the dime shows heads. a. Construct an approximate 95% con…dence interval for p1 4 p2 . b. Refer to (a). Is there reason to believe that the coins have a di¤erent probability of showing a head on a given toss? 9. Battery X has a lifetime that is normally distributed with unknown mean X and unknown variance 2 . Battery Y has a lifetime that is normally distributed with mean Y and unknown variance 2 . A random sample of 10 type X batteries showed a sample mean of 1:263 years and a standard deviation of 0:2978 years, and a random sample of eight type Y batteries showed a sample mean of 1:576 years and a standard deviation of 0:288 years. a. Construct a 99% con…dence interval for X Y . b. Refer to (a). Is there reason to believe that battery X and battery Y have a di¤erent life expectancy? c. Construct an 80% upper con…dence interval for X Y . d. Refer to (c). Is there reason to believe that the average life of Y exceeds the average life of X? e. Recall the de…nition of the pooled variance Sp2 = Show that 2 + (n 1) SY2 1) SX m+n 2 (m 2) Sp2 (m + n 2 d = 2 m+n 2 f. Using (e), derive a (1 ) 100% con…dence interval for 2 . g. Using (f) and the data in this question, …nd a 90% con…dence interval for the common variance 2 . 10. Let X1 ; X2 ; :::; Xn be iid N 0; 2 where 2 > 0 is unknown. ! n n X X 2 2 a. Give a pivotal quantity of the form g Xi ; Xi ; , where g is a i=1 i=1 function. b. Using (a), derive a (1 ) 100% con…dence interval for 2 . c. Let L1 be the length of the interval found in (b).!Find E (L1 ). n X d. Give a pivotal quantity of the form g Xi2 ; 2 , where g is a function. i=1 e. Using (d), derive a (1 ) 100% con…dence interval for 2 . f. Let L2 be the length of the interval found in (e). Find E (L2 ). g. For n = 2, and 1 = 0:95, compare E (L1 ) to E (L2 ). h. Repeat (g) for n = 10 and 1 = 0:95. Which interval seems to be longer on average? Why do you think this is? 5

Tutor Answer

Vishalc0
School: Boston College

Attached.

Question 1:

 3x 2
, 0  x 

a) Since X1 , X 2 ,..., X n are i.i.d. with the density f ( x; ) =   3
. It follows that
0, otherwise

X
X1 X 2
,
,..., n are i.i.d. with the following density:
 


  x 2
x
x 
3   , 0   1 d

f   ;1 =  f ( x; ) =    
~ U ( 0,1)

 
 0, otherwise

X 
X X
=  1 , 2 ,..., n  has the same distribution as the maximum of a random

 
 
X (n)
sample from the U ( 0,1) distribution. Therefore,
is a pivotal quantity.

Hence,

X ( n)

b) We select two numbers a and b satisfying the following conditions:


Pa 



Pa 






And since we have P  a 


 b  = 1−



X (n)
 b n −1
 b  =  nt dt = b n − a n


 a
X (n)

X (n) 

 X (n)
 b = P
 
 , it immediately implies that

 b

a 



X ( n)

 X (n) X (n) 
,

 is a (1 −  )100% confidence intervals for  , provided that 0  a, b  1 and
a 
 b
b n − a n = 1 −  . And it would be desirable to find an interval of the above form which has the

1


1

shortest possible length which is X ( n )  −  .
a b



1

1

We cannot control over X ( n) but we can pick a and b to minimize − subject to the
a b
constraint b n − a n = 1 −  . And it is straight forward that the solution is a =  1/ n and b = 1 .
Hence, the shortest (1 −  )100% confidence intervals for  is calculated to be

 X ( n) X ( n) 
shortest (1 −  )100% CI = 
,
=
a 
 b
Similarly,

X (n) 

 X ( n ) , 1/ n 
 


1 1
− is maximized when a = 0 and when b is equal to
a b

bn − a n = 1 − 
 bn − 0 = 1 − 
 bn = 1 − 
 b = (1 −  )

1/ n

Since a = 0 , we have

X (n)
a

=  . Hence, the longest (1 −  )100% confidence intervals for  is

calculated by


 X ( n) X ( n)   X ( n)
longest (1 −  )100% CI = 
,
,
=
1/ n

a   (1 −  )
 b

c) Let 1   2 be 2 distinct levels of significance where 0  1 ,  2  1 . We will use  1 for lower
confidence interval and  2 for upper confidence interval.
Hence, a (1 −  )100% lower confidence intervals for  can be found as follow

 X ( n) X ( n) 
lower (1 −  )100% CI = 
,

a 
 b
X ( n) 

 lower (1 −  )100% CI =  X ( n ) , 1/ n 
1 

where 2 numbers a and b satisfy 0  a, b  1 and b n − a n = 1 − 1 . In fact, we choose a = 11/ n
and b = 1 .
Similarly, a (1 −  )100% upper confidence intervals for  can be found as follow

 X (n) X ( n) 
upper (1 −  )100% CI = 
,

a 
 b
X (n) 

 upper (1 −  )100% CI =  X ( n ) , 1/ n 
 2 


where 2 numbers a and b satisfy 0  a, b  1 and b n − a n = 1 −  2 . In fact, we choose a =  21/ n
and b = 1 .
Hence, compared to the answer in part (b), both of the lower and upper (1 −  )100%
confidence intervals are wider than the shortest (1 −  )100% CI because of the 2 inequalities

X ( n)

 1/ n

− X ( n) 

X ( n)

11/ n

− X ( n ) and

X ( n)

 1/ n

− X ( n) 

X ( n)

 21/ n

− X ( n ) , and are less wider than the longest

(1 −  )100% CI whose right limit is  .
d) For 40% confidence interval, we can solve for  by

(1 −  )100% = 40%
40%
= 0.4
100%
  = 1 − 0.4
  = 0.6
 1− =

For sample 1, the shortest 40% confidence interval is calculated by

X ( n) 

shortest 40% CI =  X ( n ) , 1/ n 
 

0.978844113 

 shortest 40% CI = 0.978844113,

0.61/3

 shortest 40% CI =  0.979,1.161
For sample 1, the longest 40% confidence interval is calculated by

 X ( n)

longest 40% CI = 
,


1/ n

 (1 −  )

 0.978844113 
 longest 40% CI = 
,
1/3

 (1 − 0.6 )

 longest 40% CI = 1.3285,  )
For sample 2, the shortest 40% confidence interval is calculated by

X ( n) 

shortest 40% CI =  X ( n ) , 1/ n 
 

0.849384877 

 shortest 40% CI = 0.849384877,

0.61/3

 shortest 40% CI =  0.849385,1.007057 
For sample 2, the longest 40% confidence interval is calculated by

 X (n)

longest 40% CI = 
,


1/ n

 (1 −  )

 0.84938487

 longest 40% CI = 
,


1/3

 (1 − 0.6 )

 longest 40% CI = 1.152793,  )
For sample 3, the shortest 40% confidence interval is calculated by

X ( n) 

shortest 40% CI =  X ( n ) , 1/ n 
 

0.940479844 

 shortest 40% CI = 0.940479844,

0.61/3

 shortest 40% CI =  0.940479844, 1.115062
For sample 3, the longest 40% confidence interval is calculated by

 X ( n)

longest 40% CI = 
,


1/ n

 (1 −  )

 0.940479844

 longest 40% CI = 
,


1/3

 (1 − 0.6 )

 longest 40% CI = 1.276428,  )
For sample 4, the shortest 40% confidence interval is calculated by

X (n) 

shortest 40% CI =  X ( n ) , 1/ n 
 

0.90503358 

 shortest 40% CI = 0.90503358,

0.61/3

 shortest 40% CI =  0.90503358, 1.073036 
For sample 4, the longest 40% confidence interval is calculated by

 X ( n)

longest 40% CI = 
,


1/ n

 (1 −  )

 0.90503358

 longest 40% CI = 
,


1/3

 (1 − 0.6 )

 longest 40% CI = 1.22832,  )
For sample 5, the shortest 40% confidence interval is calculated by

X ( n) 

shortest 40% CI =  X ( n ) , 1/ n 
 

0.868601028 

 shortest 40% CI = 0.868601028,

0.61/3

 shortest 40% CI =  0.868601028, 1.02984
For sample 5, the longest 40% confidence interval is calculated by

 X ( n)

longest 40% CI = 
,


1/ n

 (1 −  )

 0.868601028

 longest 40% CI = 
,


1/3

 (1 − 0.6 )

 longest 40% CI = 1.178873,  )
e) It can be checked that all of the five shortest confidence intervals in part (d) contained  = 1,
and none of the five longest confidence intervals in part (d) contained  = 1. The expected
numbers show that half of the number of confidence intervals should contain  = 1.
Question 2:

 2x
 x2 
exp

 − , x  0
a) It is given that X1 , X 2 ,..., X n are i.i.d. with the density f ( x; ) =  
  
0,
x0

where   0 is unknown.
Hence, X 12 has the following density function (ignore the zero-part):

 x2  2x
 x2 
exp  −   exp  − 

   
  

( f ( x; ) ) =

2x

2

 ( f ( x; ) ) =
2

 2x2 
exp
−

2
  

4x2

 ( f ( x; ) ) = exp ( )
2 d

d

 X 12 = exp ( )
b) Summing up the result obtained in part (a) for the remaining n − 1 variables, it follows that
d

X 12 = exp ( )
n

d

  X i2 = n exp ( )
i =1



But we have

2n



2



n

d

 X i2 =
i =1

d

2n



exp ( )

exp ( ) =  22n . Hence, it immediately follows that

2

n

X

i =1

2
i

=  22n , as desired

result.

(

)

c) Let a and b be such that P a   22n  b = 1 −  . Then, we will have
n
n


 n 2

2
2
X
2
X
2
X i2 



i
i




P ( a   22n  b ) = P  a  i =1
 b  = P  i =1
   i =1

a 


 b...

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Tutor went the extra mile to help me with this essay. Citations were a bit shaky but I appreciated how well he handled APA styles and how ok he was to change them even though I didnt specify. Got a B+ which is believable and acceptable.

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