Question

15 trainees are randomly assigned to 3 groups. Tne scores at the end are below. Using the ANOVA provedure we decide the 3 sample means are significantly. different

TRUE or FALSE

A1 86, 79, 81, 70,84

A2 90. 76. 88. 82.89

A3. 82. 68 73. 71 81

Answer 1

Ans : False

We dont have sufficient evidence that sample means are significantly. different.

Let and be the means of A1, A2 and A3

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

Hence, the null hypothesis as:

i.e all sample means are equal.

and the alternative hypothesis is given by

i.e all sample 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

The mean of these sample can be calculate as

Now obtaining the grand mean ( )

= 80

The mean of each sample is obtained as

= 80

= 85

= 75

Now calculating Sum of Squares.

Total Sum of Squares (TSS)

= ( 86 - 80)2  + ( 79 - 80)2 +(81 - 80)2  +...( 71 - 80)2  + ( 81- 80)2

=698

Treatment Sum of Squares ( SST)

= 5* ( 80-80 )2+   5* ( 85 -80 )2+   5* ( 75-80 )2

=250

Error Sum of Square ( SSE)

SSE = TSS - SST

= 698-250

= 448

Obtaining Degree of Freedom

df( Treatment )= t-1

= 3-1= 2

df(error)= N-t

= 15-3

= 12

df(Total)=N-1

= 15-1

= 14

Calculating Mean Square of Error

Treatment Mean Square ( MST)

= 37.3333

Now obtaining the test statistic( F- statistic)

= 3.348214

Obtaining the critical value

from the F-table

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

Since F (calculated) = 3.348214< F ( critical value) = 3.8853,

we failed to reject null hypothesis.

We dont have sufficient evidence that sample means are significantly. different.

( Comment if needed more explantion)