Your friend believes that he has found a route to work that would make your commute faster than what it currently is under similar conditions. Suppose that data were collected for a random set of 7 days, where each difference is calculated by subtracting the time taken on the current route from the time taken on the new route. Assume that the populations are normally distributed. Your friend uses the alternative hypothesis Ha:μd<0. Suppose the test statistic t is computed as t≈−3.201, which has 6 degrees of freedom. What range contains the p-value?

The range of p-values

0.01 < p < 0.025

Step-by-step explanation:

Explanation:-

Given random sample size 'n' = 7

Assume that the populations are normally distributed

Null Hypothesis :H₀:μd=0.

Alternative Hypothesis:H₁:μd<0.

Degrees of freedom

ν = n-1 =7-1 =6

given the test statistic  t = - 3.201

we will use single tailed test in t-distribution table

The test statistic t= 3.201 is lies between the critical values is 0.01 and 0.025

The range of p-values

0.01 < p < 0.025  (check t- distribution table single tailed test)

Final -

The range of p-values

0.01 < p < 0.025

add an image

step-by-step explanation:

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!