In: Statistics and Probability
Listed below are amounts of arsenic in samples of brown rice from three different states. The amounts are in micrograms of arsenic and all samples have the same serving size.
Arkansas | 4.8 | 4.9 | 5 | 5.4 | 5.4 | 5.6 | 5.6 | 5.6 | 5.9 | 6 | 6.1 |
California | 1.5 | 3.7 | 4 | 4.5 | 4.9 | 5.3 | 5.4 | 5.4 | 5.5 | 5.6 | 5.6 |
Texas | 5.6 | 5.8 | 6.6 | 6.9 | 6.9 | 7.1 | 7.3 | 7.5 | 7.6 | 7.7 | 7.7 |
A) Use a 0.05 significance level to test the claim that the three samples are from populations with the same mean.
What is the critical value (F-value)? [Round to 4 decimal places]
B) Would you reject the null hypothesis or fail to reject the null hypothesis? [H0: μ1=μ2=μ3]
C) Given the information, what can you conclude about the claim?
Group of answer choices:
There is sufficient evidence to warrant rejection of the claim
There is not sufficient evidence to warrant rejection of the claim
D) Can we conclude that brown rice from Texas poses the greatest health problem, given that the amounts of arsenic in the samples from Texas have the highest mean?
This is an ANOVA testing for the same means. We will use F-dist to test whether all the treatments have the same mean or not.
Arkanas | Cali | texas | ||
1 | 4.8 | 1.5 | 5.6 | |
2 | 4.9 | 3.7 | 5.8 | |
3 | 5 | 4 | 6.6 | |
4 | 5.4 | 4.5 | 6.9 | |
5 | 5.4 | 4.9 | 6.9 | |
6 | 5.6 | 5.3 | 7.1 | |
7 | 5.6 | 5.4 | 7.3 | |
8 | 5.6 | 5.4 | 7.5 | |
9 | 5.9 | 5.5 | 7.6 | |
10 | 6 | 5.6 | 7.7 | |
11 | 6.1 | 5.6 | 7.7 | |
total | 60.3 | 51.4 | 76.7 | 188.4 |
Total square | 295.26 | 224.22 | 480.78 | 1000.26 |
We know
=
= 52.5273
=
Note: We have to separately sum for all the treatment.
= 29.9473
Df (Bet) = k -1 ………..where k = no. of treatments
=2
22.58
Df (res) = n – k
=32
Test Stat = MSB / MSE
ANOVA | ||||
Source of Variation | SS | df (SS /df) | MS | F |
Between Groups | 29.947 | 2 | 14.974 | 19.894 |
Within Groups | 22.58 | 30 | 0.753 | |
Total | 52.527 | 32 |
Test
Null: All the states have the same mean amount of arsenic.(μ1=μ2=μ3)
Alternative: At least one mean amount is different.
A) Use a 0.05 significance level to test the claim that the three samples are from populations with the same mean.
C.V. =
=F2,30,0.05
F-crit = 3.3158
B) Would you reject the null hypothesis or fail to reject the null hypothesis? [H0: μ1=μ2=μ3]
Since F-stat (19.894) > C.V.
We reject the null hypothesis at 5%.
C) Given the information, what can you conclude about the claim?
Since we are rejecting the there is evidence to conclude that at least one mean amount is different.
There is sufficient evidence to warrant rejection of the claim
D) Can we conclude that brown rice from Texas poses the greatest health problem, given that the amounts of arsenic in the samples from Texas have the highest mean?
Groups | Average |
Arkanas | 5.4818 |
Cali | 4.6727 |
texas | 6.9727 |
Even though the results state that at least one is different, we do not know where the difference lies between A and C or C and T or T and A. We need to conduct either Tukey's post Hoc or Fisher's LSD to find out. Therefore without testing we can not conclude that brown rice from Texas poses the greatest health problem, given that the amounts of arsenic in the samples from Texas have the highest mean.