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
See, please, suggested decision, if it is possible check it.
Answer:
ok.......
Step-by-step explanation:
Answer:
a
Step-by-step explanation:
expressions dont have answers
Answer:
x ≠ - 2, 5
Step-by-step explanation:
The restrictions occur in the denominator which cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the values that x cannot be.
solve x² - 3x - 10 = 0 ⇒ (x - 5(x + 2) = 0 ⇒ x = -2, 5
