To evaluate the effect of a treatment, a sample of n=8 is obtained from a population with a mean of μ=40 , and the treatment is administered to the individuals in the sample.
after treatment, the sample mean is found to be m=35 .

a. if the sample variance is s^2=32 , are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha=.05

b. if the sample variance is s^2=72 , are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha=.05 ?

c. comparing your answer for parts a and b, how does the variability of the scores in the sample influence the outcome of a hypothesis test?

Step-by-step explanation:

a)

Test statistic:

here test statistic lie in rejection region,that why null hypothesis fails  

so Yes, its significant.

b)

Test statistic:

c)

sample variability increases, therefore likelihood of rejecting the null hypothesis decreases.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

1) 2/5

2) zero=0/1

3) zero=0/2

4) -4/5

5) undefined=3/0