Homework 5 – Answer Set

QSC 482 – Autumn 2002

Prof. Loveday Conquest

 

Q1)

 

 

To derive raw score related to –t.10(1);9 = -1.383, note that this value implies that the raw score value of the cut-off point is 1.383 standard errors below (negative sign) the hypothesized value of 32. Therefore:

se = s/Ön = 3.6/Ö10 = 1.138

            (t.10(1);9 * se) = -1.383 * 1.138 = -1.57 à 32 - 1.57 = 30.43       

 

To determine m note that here it will be equal to the hypothesized value (mo=32) minus the MDE (d):

            d = (s/Ön)(t.10(1);9 + t.05(1);:9)

               = (3.6/Ö10)(1.383 + 1.833) = 3.66 à 32 – 3.66 = 28.34

 

2a)

Ho: m1-m2 = 0               a = .10 (two-tailed)       n1=6     n2=9     t.10(2);13 = 1.771

Ha: m1-m2 ¹ 0              

 

s2p= [(n1-1)s21 + (n2-1)s22]/(n1+n2-2)) = [(5*4.62) + (8*3.52)]/(6+9-2) = 15.6769

sxbar1-xbar2 = Ö(s2p[(1/n1)+(1/n2)] = Ö(15.6769[(1/6)+(1/9)]) = 2.0868

 

tobs = |xbar1-xbar2| - d)/ sxbar1-xbar2 = (|40.0-40.5|-0)/2.0868 = 0.2396

 

Since tobs<1.771, Do Not Reject the null and conclude that the means do not differ.

 

xbarp = (n1*xbar1 + n2*xbar2)/n1+n2 = ((6*40)+(9*40.5))/15 = 40.3

           

(1-a) 100% CI = xbarp - ta(2);n1+n2-2Ö(s2p/(n1+n2))£ m £ xbarp + ta(2);n1+n2-2Ö(s2p/(n1+n2))

                        = 40.3 - 1.8105 £ m £ 40.3 + 1.8105

                        = (38.49, 42.11)

 

2b)

Step 1:

            n ³ (2s2/d2)(ta,¥ + tb(1);¥)2 = ((2*42)/.52)(1.6449 + 0.6745) 2 =

            = (128)(5.3796) = 688.59 à 689

 

689*2 = 1378 à df = n1+n2-2 = 689+689-2 = 1376

 

            Step 2:

Since interpolating with an upper bound of ¥ is not really possible, instead just try rounding down degrees of freedom to next smallest value (i.e., df=1000) and see what happens…

 

=((2*42)/.52)(1.646 + 0.675) 2 = 689.541 à 690

 

690*2 = 1380 à df = n1+n2-2 = 1378 à STOP

 

Total sample size is 690 for each sample for a grand total of 1380.

 

2c)

            tb(1);n1+n2-2= d/Ö(sp2(1/n1)+(1/n2))- ta;n1+n2-2

            tb(1);13= [1/Ö(16((1/6)+(1/9))]- 1.771 = -1.2967

 

            .10<P(tb(1);13£-1.2967)<.25

 

2d)

            Ho: s21=s22

Ha: s21¹s22

 

F5;8= 3.69

 

            F = 4.62/3.52 = 1.7273

 

Since 1.7273<3.69 we Do Not Reject the null and conclude that the variances are not different.

 

3a)

            Test D = Pop1-Pop2

Ho: D£0

Ha: D >0                       Ua,n1,n2 = U.10(1);7;9 = 45

 

 

trace

1.01

1.17

1.27

1.45

1.54

1.71

1.71

1.74

1.79

1.81

1.91

2.0

2.0

2.11

2.3

2

2

2

2

2

2

2

2

1

1

1

1

1

1

1

2

1

2

3

4

5

6

7.5

7.5

9

10

11

12

13.5

13.5

15

16

 

R1 = 9+10+11+12+13.5+13.5+15 = 84

R2 = 1+2+3+4+5+6+7.5+7.5+16 = 52

 

Check: N(N+1)/2 = 16*17/2 = 136 = R1+R2

 

U1 = n1*n2+(n1(n1+1))/2) – R1 = (7*9)+((7*8)/2)-84 = 63 + 28 – 84 = 7 ß Not needed!!!

Note that since this is a one-sided Ha, calculating U1 is unnecessary. We only need focus on R2 since it will be smaller in expected value under Ha (see pgs. 1-10 to 10-13 of class notes). This saves you a little bit of work…

 

U2 = n1*n2+(n2(n2+1))/2) – R2 = (7*9)+((9*10)/2)-52 = 63 + 45 – 52 = 56

 

Umax = 56

 

Since Umax>45 Reject the null and conclude that the concentration of Am-241 in population 1 does exceed that of population 2.