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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

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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.

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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.

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