I would guess the answer to be Tension, if not for the extra 'X'....
or
Extension, if you add another 'E'...
Answer:
Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0
Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0
Here to test Fertilizer A height is greater than Fertilizer B
Two Sample T Test:
t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}
Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}
S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154
t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524
P value for Test Statistic of P(3.3524,26) = 0.0012
df = n1+n2-2 = 26
Critical value of P : t_{0.025,26}=2.05553
We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.
Answer:
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Step-by-step explanation:
Given
Required
Determine the hourly rate
First, we need to determine the ice cream sold
Next, we calculate the hourly rate:
<em /><em> -- Approximated</em>
<span>45% -------------------------------chance that the land has oil
then
55</span>%--------------------------------chance that the land not have oil
<span>80% -------------------accuracy rate of indicating oil in the soil. (if land does have oil)
20</span>% -------------------accuracy rate of indicating not oil in the soil. (if land does have oil)
20% ---------------------accuracy rate of indicating oil in the soil. (if land has not oil)
80% ---------------------accuracy rate of indicating not oil in the soil. (if land has not oil)
that the land has no oil----------------55%
the test shows that it has oil-------- 20%
then 0.55*0.20=0.11=11%
the probability that the land has no oil and the test shows that it has oil is 11%
