In: Statistics and Probability
Using the partial ANOVA Table found in question d-2, please answer the following questions. Use the 0.05 significance level.
How many treatments are there?
What is the total sample size?
What is the critical value of F? (Round your answer to 2 decimal places.)
d-1. Write out the null and alternate hypothesis.
d-2. From the information given above, fill in the blanks in the ANOVA table below. Calculate the value of F also. (Round the values to the nearest whole number.)
source | Sum of Squares | df | Mean Square | F |
treatments | 2 | |||
error | 20 | |||
total | 500 | 11 |
What is your decision regarding the null hypothesis for the 0.05 significance level?
Answer:
Given Data
a) How many treatments are there.
The number of treatments is determined below :
From the given information d - 2 , the total number of treatments are 2 and the total degrees of freedom is 11.
The required total number of treatments is,
The total number of treatments is 3.
Explanation :
The total number of treatments is obtained by equating the treatment degrees of freedom with 2.
b) What is the total sample size .
The total sample size is obtained below:
From the given information d - 2 , the degrees of freedom for the total sum of squares equals 11.
The required total number of treatments is ,
The total sample size is 12 .
c ) What is the critical value of F.
The critical value for a right tail F - test at = 0.05 is obtained as shown below :
The critical value , is calculated based on numerator degrees of freedom ( k - 1 ) and denominator degrees of freedom ( n - k)
The denominator degrees of freedom is obtained below :
The numerator degrees of freedom is 2 and the denominator degrees of freedom is 9 .The level of significance is = 0.05.
From 'F' distribution table , = 0.05 ', locate the 'column 2 ' in numerator degrees of freedom . Also , locate the ' row 9 ' In denominator degrees of freedom .
The critical value for a right tail F-test with = 0.05 is ,
The critical value F is 3.01.
Explanation :
The critical value for a right tail F - test is obtained by locating the numerator degrees of freedom and denominator degrees of freedom in the F distribution table at = 0.05 significance level.
(d-1)
The null and alternative hypothesis are stated as follows:
Null hypothesis:
Alternative hypothesis :
The null hypothesis is
and the alternative hypothesis is .
( d - 2)
The F - table was obtained below :
From the information , it is clear that
SSE = 20 ,
SST = 500,
,
,
,
The sum of squares of error is obtained below :
SSE = SST - SSTR
20 = 500 - SSTR
SSTR = 500 - 20
= 480
The value of Mean sum of squares due to treatment is obtained below :
= 240
The value of Mean sum of squares due to error is obtained below :
= 2.222
The value of F - statistic is obtained below :
= 108.0011
The complete F - table is determined below :
Source of variation | Sum of squares | Degrees of freedom | Mean square | F |
Treatments | 480 | 2 | 240 | 108 |
Error | 20 | 9 | 2.2222 | |
Total | 500 | 11 |
part d - 2
The completed F table is
Source of variation | Sum of squares | Degrees of freedom | Mean square | F |
Treatments | 480 | 2 | 240 | 108 |
Error | 20 | 9 | 2.2222 | |
Total | 500 | 11 |
Explanation :
The complete F - table is obtained by finding the sums of squares due to error , the mean squares due to treatment , error and the F - statistic by using the concept of ANOVA.
e) What is your decision regarding the null hypothesis for the 0.05 significance level.
The decision regarding null hypothesis for the 0.05 significance level is stated below :
Use the significance level = 0.05
The value of the test statistic is 108 and the critical value is 3.01.
That is ,
Therefore , by the rejection rule , reject the null hypothesis
Rehect , there is significant difference between the three treatment means.
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