Hwk9Soln
-
Upload
aditya-goyal -
Category
Documents
-
view
218 -
download
0
Transcript of Hwk9Soln
-
8/2/2019 Hwk9Soln
1/5
INTRODUCTION TO PROBABILITY AND STATISTICS II - SPRING
2010
Assignment 9 - SOLUTIONS
Problem 1
a) H0 : = 1b) Ha : = 1c)
Q =21S
22
22S21
=S22/
22
S21/21
=
(m
1)S22/
22(m
1)
(n 1)S21/21(n 1)=
[(m 1)S22/22] /(m 1)[(n 1)S21/21] /(n 1)
But,(n1)S2
1
21
2n1 and (m1)S2
2
22
2m1.Hence, Q Fm1,n1 by the definition of the F distribution.
d) Our pivot is Q =21S22
22S21
Fm1,n1. So we set
1 = PrFm1,n1,1/2 21S
22
22S21 Fm1,n1,/2
= Pr
S21S22
Fm1,n1,1/2 21
22 S
21
S22Fm1,n1,/2
Hence a 1 CI for = 21/22 isS21S22
Fm1,n1,1/2,S21S22
Fm1,n1,/2
e) In the deer example, n = 40, m = 40, S21 = 11402, S22 = 963
2. So our CI is11402
9632F39,39,0.975,
11402
9632F39,39,0.025
11402
96321
1.88,
11402
96321.88
(0.75, 2.63)
f) Becasue the C.I. includes the value 1, H indicates that the ratio of the variancesmay be equal. We have no evidence that the assumption we made was violated.
g) We reject if s2
2/s2
1 > k1 or s2
2/s2
1 < k2.
Need
Pr
S22S21
> k1
H0
= /2
1
-
8/2/2019 Hwk9Soln
2/5
and
Pr
S22S21
< k2
H0
= /2.
Now,
Pr
S22S21
> k1
2122 = 1
= Pr
S22S21
2122
> k1
= Pr(F > k1) where F Fm1,n1.
Hence, k1 = Fm1,n1,/2.
Similarly
Pr
S2
2S21
< k2 2122 = 1 = PrS
2
2S21
2
122
< k1
= Pr(F < k2) where F Fm1.n1.Hence, k2 = Fm1,n1,1/2.
So, we reject ifS22
S21
> Fm1,n1,/2 orS22
S21
< Fm1,n1,1/2.
Because X Fm,n 1X Fn.m, we can rewrite this as reject iflargest sample var
smallest sample var> FV1,V2,,
where V1 =df for largest var and V2 =df for smallest var.
h) In the deer example,S22S21
=9632
11402= 0.713
F39,39,0.025 1.88F39,39,0.975 1
1.88 0.53
Since 0.713 1.88 and 0.713 0.53, we do not reject the null hypothesis that = 1.
We dont have evidence that the variances in the two groups are different.
i) Yes.The 95% C.I. should yield the same conclusion as a two-sided hypothesis test sincethey are based on the same pivot.
Note that although this test indicates that our assumption of equal variance wasprobably valid, it itself assumes that the data are normally distributed, which wehad not assumed in our analysis of the distances! Its really hard to get rid of allassumptions.
2
-
8/2/2019 Hwk9Soln
3/5
Problem 2
a)
1. The parameter of interest is p, the proportion of overweight American children.2. H0 : p = 0.15
3. Ha : p < 0.15
4. Because we have a large sample, we can invoke the CLT and use the test statistic
Z =pp0
p(1 p)/n =0.13 0.15
0.13(0.84)/100=0.020.033
= 0.59
5. We reject H0 if Z 0 (we expect it to take longer for TT list)4. First, we compute
s2p =(n 1)S21 + (m 1)S22
n + m 2 =1.94(41) + 1.92(41)
42 + 42 2 = 1.93
Thus, our test statistic is
Z =1 2 0sp
1n
+ 1n
=6.59 6.34
1.93
142
+ 142
=0.25
0.3031= 0.82
5. We reject if Z > Z = 1.64
6. 0.82 < 1.96, we dont reject. There is not enough evidence to say that it takesmore time to read the TT lists than the control lists.
d) The independence between the Xi and Yi, because the same people read both typesof lists, so Xi and Yi are dependent.
Some people also mentioned that the assumption that the variance is equal for thetwo groups could be violated. This is not as problematic because we dont know forsure that it is violated and it is possible to test for it and change our test if needed.
e) Because there is probably a large variability between students and their skills at
reading lists and identifying non words. To make sure that the difference observed iscaused by the treatment, they give each type of list to each student.
f) H0 : 0 = 0. Ha : 0 > 0.
g) Our test statistic is
t =D 0SD/
n
0.25
0.78/42 2.07
Since t = 2.07 > t41,0.05 Z0.05 1.96, we reject H0. There is evidence that thereading time is longer for TT lists.
h) Yes. We both get a t-statistic of 2.07 to compare to a t41.
i) With the test for the equality of means for two independent samples of (c), we didnot reject H0, but we did using the paired t-test in (g).
This is because the variability of scores between the students within a group was toolarge to allow us to discern the difference between the two groups.
4
-
8/2/2019 Hwk9Soln
5/5