The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let µ denote the true average reflectometer reading for a new type of paint under consideration. A test of H0: µ = 20 versus Ha: µ > 20 will be based on a random sample of size n from a normal population distribution. What conclusion is appropriate in each of the following situations? (Round your P-values to three decimal places.)

(a) n = 16, t = 3.3, a = 0.05
P-value =
(b) n = 8, t = 1.8, a = 0.01
P-value =
(c) n = 26,
(d) t = -0.6

(a) P-value = 0.001

Conclusion: The average reflectometer reading for the new type of paint under consideration is greater than 20.

(b) P-value = 0.036

Conclusion: The average reflectometer reading for the new paint under consideration is 20

Step-by-step explanation:

The test is a one-tailed test because the alternate hypothesis is expressed using greater than.

(a) n = 16, t = 3.3, a = 0.05

Cumulative area of the test statistic t is 0.9995

P-value = 1 - 0.9995 = 0.0005 = 0.001 (to 3 decimal place)

Conclusion:

Reject H0 because the P-value 0.001 is less than the significance level 0.05.

(b) n = 8, t = 1.8, a = 0.01

Cumulative area of the test statistic t is 0.9641

P-value = 1 - 0.9641 = 0.0359 = 0.036 (to 3 decimal places)

Conclusion:

Fail to reject H0 because the P-value 0.036 is greater than the significance level 0.01

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!