Reconsider the paint-drying situation of Example 8.2, in which drying time for a test specimen is normally distributed with σ = 9. The hypotheses H0 : µ = 75 versus Ha : µ < 75 are to be tested using a random sample of n = 25 observations.

(a) How many standard deviations (of X¯) below the null value is ¯x = 72.3?

(b) If ¯x = 72.3, what is the conclusion using α = 0.002?

(c) For the test procedure with α = 0.002, what is β(70)?

(d) If the test procedure with α = 0.002 is used, what n is necessary to ensure that β(70) = 0.01? (e) If a level 0.01 test is used when n = 100, what is the probability of a type I error when µ = 76.

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

0.6

39mph

step-by-step explanation: