At a canning facility, a technician is testing a machine that is supposed to deliver 250 milliliters of product. The technician tests 44 samples and determines the volume of each sample. The 44 samples have a mean volume of 251.6 mL. The machine is out of calibration when the average volume it dispenses differs significantly from 250 mL. The technician wants to perform a hypothesis test to determine whether the machine is out of calibration. Assume σ = 5.4 is known. Compute the value of the test statistic.

The value of the test statistic is

Step-by-step explanation:

The null hypothesis is:

The alternate hypotesis is:

Our test statistic is:

In which X is the sample mean, is the null hypothesis value, is the standard deviation and n is the size of the sample.

In this problem:

So

The value of the test statistic is

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