Question

1.  A researcher compares emotionality in different breeds of 4-week-old dogs.  Each pup is handled for 25 seconds and the duration of time spend struggling while being handled is measured for each pup.  The data are as follows:

German Shepherd          Cocker Spaniel          American Pit Bull

             4                                   2                                   8

             3                                   0                                   4                     N = 12

             5                                   2                                   6                     G = 44

             4                                   0                                   6                  ∑X² = 226

         T = 16                           T = 4                            T = 24

        SS = 2                           SS = 4                          SS = 8

Do differences in emotionality exist among the dog breeds?  If significant, use a Tukey post-hoc test to compare the breeds.  Use the .01 level of significance.

2.  A clinical psychologist has noted that autistic children seem to respond to treatment better if they are in a familiar environment.  To evaluate the influence of environment, the psychologist selects a group of 15 autistic children who are currently in treatment and randomly divides them into three groups.  One group continues to receive treatment in the clinic as usual.  For the second group, treatment sessions are conducted entirely in the child’s home.  The third group gets half of the treatment in the clinic and half at home.  After 6 weeks, the psychologist evaluates the progress for each child.  The data are as follows:

Clinic               Home                Both

    1                       2                       4

    1                       7                       1          N = 15

    5                       2                       2          G = 45

    2                       4                       2       ∑X² = 191

    1                       5                       6

T = 10              T = 20              T = 15

SS = 12           SS = 18            SS = 16

Do the data indicate any significant differences between the three settings?  Test at the .05 level of significance.  If significant, use a Scheffé post-hoc test to compare the settings.

Answer 1

Q1)

  There is differences in emotionality exist among the dog breeds

There is significantly difference emotionality between  Cocker Spaniel and American Pit Bull breed.

##########################Explanation ##############

Let the average of the duration of time spend struggling while being handled  different breeds be given as

The null hypothesis for an anova always assumes the population means are equal.

Hence, the null hypothesis as:

i.e all treatment means are equal.

and the alternative hypothesis is given by

i.e all treatment means are not equal.

Let X1 ,X2 and X3  be the the duration of time spend struggling while being handled German Shepherd , Cocker Spaniel and   American Pit Bull

Let Xij be the observations.

i = 1,2 ...r. r is total number of rows

j= 1,2,...t t is total number of treatment.

Here

r=4

t=3

N= Total observation

= r*t

=12

The mean of these treatment can be calculate as

Now obtaining the grand mean ( )

= 3.667

Now calculating Sum of Squares.

Total Sum of Squares (TSS)

= (4- 3.667)2+ (3- 3.667)2+...+(6- 3.667)2

= 64.6667

Treatment Sum of Squares ( SST)

= 4*(4-3.667)2+4*(1-3.667)2+4*(6-3.667)2+

= 50.6667

Error Sum of Square ( SSE)

SSE = TSS - SST

= 64.6667-50.6667

Obtaining Degree of Freedom

df( Treatment )= t-1

= 2

df(error)= N-t

= 12-3

= 9

df(Total)=N-1

= 11

Calculating Mean Square of Error

Treatment Mean Square ( MST)

= 50.6667/2

= 25.3333

=  64.6667/11

= 5.87878

= 14/9

= 1.5556

Now obtaining the test statistic( F- statistic)

= 1.5556/25.3333

= 16.2857

The critical value is given as

At 1% level of significance

= F0.01,2,9

From table

F0.0,2,9= 8.02

Decision Rule:

Reject H0 null hypothesis if F (calculated) > F(critical value)

Here F (calculated) = 16.2857 >   F(critical value) = 8.02

We reject the null hypothesis.

Hence there is differences in emotionality exist among the dog breeds

Tukey post-hoc test

.The Tukey's HSD is given as

Where q =constant obtained from the studentised range q table

MSE =Mean square error or  Mean square within

nj= Number in each category

Obtaining the q from the studentised range q table

At 1% level of significance

= 8.02

Now comparing each mean difference to~ 5.

.

Hence there is significant differeance between X2 and X3.

There is significantly difference emotionality between  Cocker Spaniel and American Pit Bull breed.

2)

There is no significant differences between the three settings Clinic ,Home  and Both.

####################Explanation ############

Let and be the mean of treatment received in Clinic ,Home and Both

The null hypothesis for an anova always assumes the population means are equal.

Hence, the null hypothesis as:

i.e all treatment means are equal.

and the alternative hypothesis is given by

i.e all treatment means are not equal.

Let Xij be the observations.

i = 1,2 ...r. r is total number of rows

j= 1,2,...t t is total number of treatment.

Here

r=5

t=3

N= Total observation

= r*t

=15

Now obtaining the grand mean ( )

= 3

= 2

=4

= 3

Now calculating Sum of Squares.

Total Sum of Squares (TSS)

= (1-3)2+(1-3)2+...(2-3)2+(6-3)2

= 56

Treatment Sum of Squares ( SST)

  

= 5*(2-3)2+5*(4-3)2+5*(3-3)2

= 10

Error Sum of Square ( SSE)

SSE = TSS - SST

= 56-40

Obtaining Degree of Freedom

df( Treatment )= t-1

= 2

df(error)= N-t

= 12

df(Total)=N-1

= 14

Calculating Mean Square of Error

Treatment Mean Square ( MST)

= 10/2

= 5

= 56/14

4

= 46/12

= 3.8333

Now obtaining the test statistic( F- statistic)

= 5/3.8333

= 1.304348

The critical value is given as

F0.05,2,12

From table

F0.05,2,12

= 3.8853

Decision Rule:

Reject H0 null hypothesis if F (calculated) > F(critical value)

Here F (calculated) = 1.304348 <  F(critical value) =3.8853

We failed to reject null hypothesis .

Hence there is no significant differences between the three settings Clinic ,Home  and Both