Answer:
Check the explanation
Step-by-step explanation:
Let
and
be sample means of white and Jesse denotes are two random variables.
Given that both samples are having normally distributed.
Assume
having with mean
and
having mean ![\mu_{2}](https://tex.z-dn.net/?f=%5Cmu_%7B2%7D)
Also we have given the variance is constant
A)
We can test hypothesis as
![H0: \mu_{1} = \mu_{2}H1: \mu_{1} > \mu_{2}](https://tex.z-dn.net/?f=H0%3A%20%5Cmu_%7B1%7D%20%3D%20%20%5Cmu_%7B2%7DH1%3A%20%5Cmu_%7B1%7D%20%3E%20%5Cmu_%7B2%7D)
For this problem
Test statistic is
![T=\frac{\overline {x}-\overline {y}}{s\sqrt{\frac {1}{n1} +\frac{1}{n2}}}](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B%5Coverline%20%7Bx%7D-%5Coverline%20%7By%7D%7D%7Bs%5Csqrt%7B%5Cfrac%20%7B1%7D%7Bn1%7D%20%2B%5Cfrac%7B1%7D%7Bn2%7D%7D%7D)
Where
![s=\sqrt{\frac{(n1-1)*s1^{2}+(n2-1)*s2^{2}}{n1+n2-2}}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B%5Cfrac%7B%28n1-1%29%2As1%5E%7B2%7D%2B%28n2-1%29%2As2%5E%7B2%7D%7D%7Bn1%2Bn2-2%7D%7D)
We have given all information for samples
By calculations we get
s=2.41
T=2.52
Here test statistic is having t-distribution with df=(10+7-2)=15
So p-value is P(t15>2.52)=0.012
Here significance level is 0.05
Since p-value is <0.05 we are rejecting null hypothesis at 95% confidence.
We can conclude that White has significant higher mean than Jesse. This claim we can made at 95% confidence.
Answer:
1. D
Step-by-step explanation:
If you calculated it you would get 20.25 so it is irrational
Answer:
Quadrant III ( C )
Quadrant IV ( D )
Step-by-step explanation:
Ordered pair ( a, b )
gives a point P on the coordinate plane
( a, b ) = ( x, y )
given : b = negative ( y-axis )
a ≠ 0 ( i.e. a = negative or positive ) ( x -axis )
Therefore point P is located in the : Third and fourth quadrants
Answer:
D is the answer. The graph was reflected across the y-axis, and shifted 1 unit downwards.
Step-by-step explanation:
Count the number of spaces to D, then to D'. They are the same (besides one being negative). A'B'C'D' was then shifted down 1 unit, which is why it is lower than shape ABCD.