A sample of 173 students using Method 1 produces a testing average of 88. A sample of 127 students using Method 2 produces a testing average of 53.5. Assume the standard deviation is known to be 12.22 for Method 1 and 13.59 for Method 2. Determine the 95% confidence interval for the true difference between testing averages for students Method 2.

43 < μ₁ - μ₂  < 49

Step-by-step explanation:

Sample 1:

Mean      x₁    =  88

Sample size    n₁  = 173

Sample standard deviation    s₁  =  12,22

Sample 2:

Mean      x₂    =  53,5

Sample size    n₂  = 127

Sample standard deviation    s₂  =  13,59

CI   95 %        α  =  5%    α = 0,05    α/2  = 0,025

We need to find:

(  x₁  -  x₂  ) - tα/2,v *√ (s₁²/n₁ ) + (s₂²/n₂) < μ₁ - μ₂  < (  x₁  -  x₂  ) + tα/2,v *√ (s₁²/n₁ ) + (s₂²/n₂)

v = degree of fredom    

v  = [ ( s₁²/n₁ + s₂²/n₂)² / (s₁²/n₁)² /n₁-1  + (s₂²/n₂)²/n₂-1

v =  [ (0,86 + 1,45 ) / 0,0043 +  0,017

v = 2,31 / 0,0213

v = 108

Then  t 0,025, 108  from t table is:  We will take v = 100

t = 1,984

Now

√ (s₁²/n₁ ) + (s₂²/n₂)  =  √ 0,86  + 1,43

√ 2,3   = 1,51

Then: CI:

(173 - 127 ) -  1,984*1,51    < μ₁ - μ₂  < ( 173 - 127 )  +  1,984*1,51

46 - 3  < μ₁ - μ₂  < 46 + 3

43 < μ₁ - μ₂  < 49

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42

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!