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

what are the things "represented below"?

step-by-step explanation:

nobody can if you dont give the choices

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3