Answer:
There is no difference as per statistical evidence.
Step-by-step explanation:
We calculate t statistic from the formula
t =difference in means/Std error of difference
Here n1 = n2
t = (x bar - y bar)/sq rt of s1^2+s2^2
Let treatment I =X = 34 41 38 29
Treatment II Y = 39 48 35 36
X Y
Mean 35.50 39.50
Variance 81.00 105.00
H0: x bar = y bar
Ha: x bar not equal to y bar
(Two tailed test at 0.05 significant level)
N1 = 4 and N2 = 4
df=N1+N2-2 = 6
s1^2 = 81/3 =27 and s2^2 = 105/3 = 35
Std error for difference =
t = -1.02
p =0.348834
p>0.05
Since p value >alpha we accept null hypothesis.
Hence there is statistical evidence to show that there is no difference in the mean level of scores.
Answer:
A) 9
Step-by-step explanation:
f(2) is basically asking what the y-coordinate is at x = 2
If you trace x = 2, you see it intersects f(x) at (2, 9).
5 out of 20 cars are red
Therefore 1 out of 4 cars is red
1/4 of cars are red
1/4 x 100 = 25 red cars
Answer:
(foh)(6) = -29
Step-by-step explanation:
(f o h)(6) is f(h(x)) when x = 6.
We have that:
f(x) = 4x - 1
h(x) = -x-1
f(h(x)) = 4(-x-1) - 1 = -4x - 4 - 1 = -4x - 5
(foh)(6) = -4(6) - 5 = -29
(foh)(6) = -29
Answer:
a. 54.05 Mpbs.
b. 2.745... standard deviations.
c. The z-score is 2.745....
d. The carrier's highest data speed is significantly high.
Step-by-step explanation:
a. The difference between the highest measured data speed and the mean is 72.6 - 18.55 = 54.05 Mbps.
b. The amount of standard deviations of 54.05 Mbps is equal to this value divided by the standard deviations, so we yield 
standard deviations.
c. The z-score is equal to the difference between the mean and a data point in standard deviations, so the z-score is 2.745....
d. 2.745... is not between -2 and 2, so the carrier's highest data speed is not insignificant - so it's significantly high.
