QSCI 482 HW 3 DUE FRIDAY, OCTOBER 19, 2002 [Exam 1 is on OCT 23!]

 

1.  Refer back to the example from Topic 1, "The Language of Singles

Bars".  We already showed that the data do NOT conform to the null

hypothesis of uniformity (we rejected that null hypothesis at the .05

level of significance).  Now use statistical partitioning (subdividing)

to show that while two of the categories "look like each other"

(uniformity between those two categories), they are different from the

third category in terms of frequency of counts. Be sure to include the

final step--showing that your chisquare test statistics from the

subdivisions add up [approximately] to the original, overall test

statistic; and that the df add up also. 

 

Type                            Observed                    Expected        

Compliments              85                                106.666

Declarations               122                              106.666

Questions                   113                              106.666

 

Partitioning     fobs                   fexp            (fobs-fexp)2/fobs

D                     122                  117.5               0.172  

Q                     113                  117.5               0.172

Total               235                  235

                                                            c2obs = 0.344

                                                            c2(0.05,1 df) = 3.841

…There isn’t a statistically significant difference between declarations and compliments at the 0.05 level of significance with 1 degree of freedom.

 

D+Q                235                  213.333           2.201

C                     85                    106.666           4.401

Total               320                  320

                                                            c2obs = 6.602

                                                            c2(0.05,1 df) = 3.841

…There is a significant difference between compliments and the other two types at the 0.05 level of significance with 1 degree of freedom.

 

Note:  The two observed c2 values above, 0.344 and 6.602 add up to 6.946, which is pretty close to the original observed statistic from topic 1, 6.98, and the degrees of freedom above add up to 2, as in the original problem.

 

 

2. A scientist has run a bioassay experiment, where each organism (say,

an amphipod) is exposed to a toxicant for a specified period of time

(96-hour bioassays are quite common). At the end of the 96 hours, the

status (alive or dead) of each organism is noted.  There are 50

organisms in each group.  The data are as follows:

                                                Expected

TREATMENT:  A     B     C     D     Total       A             B             C             D             Total  

 

Dead        01    10    15    18    44          11           11           11           11           44   

Alive       49    40    35    32    156         39           39           39           39           156

Total       50    50    50    50    200

 

At the .05 level of significance, test the null hypothesis of "no effect

of treatment type on mortality" for the 4 treatments.  Then, go through

the process of partitioning (subdivision) to get more information from

the original test result and to see which of the treatment groups can

be combined and which seem different. 

 

H0:  There is no difference in the effects of the four treatment types on mortality.

Ha:  There is some difference in the effects of the treatment types.

 

Ho:  p11 = p12 = p13 = p14

Ha:  There is some inequality involved

 

a = 0.05          df = (r-1)(c-1) = (2-1)(4-1) = 3

 

c2obs = (1-11)2/11 + (10-11)2/11 + (15-11)2/11 + (18-11)2/11 + (49-39)2/39

            + (40-39)2/39 + (35-39)2/39 + (32–39)2/39

         = 9.091 + .091 + 1.455 + 4.455 + 2.564 + 0.026 + 0.410 + 1.256

         = 19.348

 

c2(0.05,3) = 7.815

 

Reject Ho:  Conclude that there is a statistical difference in the effects of the treatment types

 

Deviations                                                                  Stdz

Treatment:     A         B         C         D                     A         B         C         D

Dead               -10       -1         +4        +7                    -3.02    -0.30    1.21     2.11

Alive               +10      +1        -4         -7                     1.60     0.16     -0.64    -1.12

 

Let’s test to see whether B, C and D might be combined

 

Obs:    B         C         D         Total               Exp      B         C         D         Total

Dead  10        15        18        43                    Dead   14.3     14.3     14.3     43

Alive   40        35        32        107                  Alive   35.67   35.67   35.67   107

Total   50        50        50        150                  Total   50        50        50        150

 

c2obs = 1.293+0.0343+0.09573+0.5256+0.01258+0.3776

            = 3.2004

 

c2(0.05,2) = 5.991

 

The statistically similar results of B, C and D indicate that they can be combined as one group

 So, let’s compare A as a single group against B,C and D as a pooled group.

 

Obs:    A         B,C&D           Total               Exp:    A         B,C&D           Total

Dead   01        43                    44                                11        33                    44

Alive   49        107                  156                              39        117                  156

Total   50        150                  200                              50        150                  200

 

c2obs = 3.0303+9.091+0.8547+2.5641

         = 15.5401

 

c2(0.05,1) = 3.84

 

 

We do reject Ho here, and conclude that A is distinct from B,C,D.   And, c2obs of 15.54 + c2obs of 3.20 = 18.74, which is roughly equal to our original 19.348.  The df add up as well (1+2 = 3).

 

     

3. The population of body weights for a small mammal is normally

distributed with a mean of 64.0 g [grams] and a standard deviation of 12.2 g.

 

a.  What is the probability that an individual drawn at random from this

population has a weight of at least 60 g?

 

m = 64.0   s = 12.2   x = 60

 

Z = x-m / s  = 60-64 / 12.2  =  -0.328            

 

Zval for 0.33 = 0.3707, therefore 0.3707 probability that an individual drawn at random would have a weight less than 60 grams, so there is an  0.6293 probability of an individual weighing 60 or more grams being drawn randomly.

 

b.  What proportion of this population has a weight between 58 and 68 g?

 

Z58 = 58-64 / 12.2  =  -0.492   From table B.2 this has a prob. val. of 0.3121 (prob. of value falling in the left tail of the distribution)

Z68 = 68-64 / 12.2 = 0.329  From table B.2 this has a prob. value of 0.3707 (prob. of value falling in the right tail of the distribution)

Therefore, 1.0000-0.3121-0.3707 = the probability of a value falling between these two amounts in the distribution.

Pr (58<x<68) = 0.3172           31.72% of the population weighs between 58 and 68g.

 

c.  If a random sample of size 12 is drawn from this population, what is

the probability that the sample average will be between 59 and 65 g?

 

sXbar = sqrt(s/n) = sqrt (12.2/sqrt 12)  =  3.52

Pr (Xbar≤59) = Pr [ Z ≤ (59-64.0)/3.52 ] = Pr(Z≤-1.42) = 0.0778

Pr (Xbar≥65) = Pr [Z ≥ (65-64.0)/3.52 ] = Pr(Z≥.2841) = 0.3897

Pr (59≤Xbar≤65) = 1.000-0.0778-0.3897 = 0.5325

 

d.  How large a sample size would one have to take to end up with a

standard error of the mean no greater than 1.0 g?

 

sxbar = sqrt(s2/n)        sxbar2 = s2/n                1.0 = 148.84/n             n must be at least 149