Approximately 68% of a normal distribution lies within one standard deviation of the mean, so this corresponds to students with scores between (57.5 - 6.5, 57.5 + 6.5) = (51, 64)
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:
C
Step-by-step explanation:
In this question, we are interested in calculating the z-score of a company employee.
Mathematically;
z-score = (x- mean)/SD
where in this case;
x is the value given which turns out to be the annual salary of the employee = 28,000
Mean = 34,000
standard deviation = 4,000
Plugging these values into the equation above, we have;
z-score = (28000-34000)/4000 = -6000/4000 = -1.5
