Answer:
The correct answer is No.
Choosing between the critical value method or the P-value method does not affect one's conclusion because both methods look at the probability of the test statistic's and its level of significance
.
Given the methodology utilized by both methods, they usually arrive at the same conclusion.
Cheers!
Answer:
The variance in weight is statistically the same among Javier's and Linda's rats
The null hypothesis will be accepted because the P-value (0.53 ) > ∝ ( level of significance )
Step-by-step explanation:
considering the null hypothesis that there is no difference between the weights of the rats, we will test the weight gain of the rats at 10% significance level with the use of Ti-83 calculator
The results from the One- way ANOVA ( Numerator )
with the use of Ti-83 calculator
F = .66853
p = .53054
Factor
df = 2 ( degree of freedom )
SS = 23.212
MS = 11.606
Results from One-way Anova ( denominator )
Ms = 11.606
Error
df = 12 ( degree of freedom )
SS = 208.324
MS = 17.3603
Sxp = 4.16657
where : test statistic = 0.6685
p-value = 0.53
level of significance ( ∝ ) = 0.10
The null hypothesis will be accepted because the P-value (0.53 ) > ∝
where Null hypothesis H0 = ∪1 = ∪2 = ∪3
hence The variance in weight is statistically the same among Javier's and Linda's rats
Answer:
Step-by-step explanation:
2n+15>3
m • (g4c - 3)
(1):g4 was replaced by g^4.
Pulling out like terms:
2.1:Pull out like factors:
g4cm - 3m = m • (g4c - 3)
Trying to factor as a Difference of Squares :
2.2:Factoring: g4c - 3