Aresearcher would like to test whether a new teaching method enhances the concept understanding. in a paired sample t test, a random sample of 20 students were given a pre-test before a new method was introduced. after the new method was introduced , a post test was given. the differences of pre-test scores and post test scores ( d = post test score - pre test score) were recorded for all students. the sample mean difference and the standard deviation of differences for the 20 students were 2.82 and 3.7 respectively. what is the value of the t test statistic (for two decimal places) for the paired sample test?

t= 3.41

Step-by-step explanation:

The t test statistic for the paired sample test is given by :-

, where n = sample size.

= sample mean difference for n

= sample standard deviation of differences for n.

Given : The sample mean difference and the standard deviation of differences for the 20 students were 2.82 and 3.7 respectively.

As per given , we have

n=20

Then , the test statistic would be :-

Hence, the  value of the t test statistic for the paired sample test : t= 3.41

what?

step-by-step explanation:

what do you need with. i love quadrants. can you tell me ?

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12