Statistics Help PLEASE!!!!!

Twenty students randomly assigned to an experimental group receive an
instructional program; 30 in a control group do not. After 6 months, both groups
are tested on their knowledge. The experimental group has a mean of 38 on the
test (with an estimated population standard deviation of 3); the control group
has a mean of 35 (with an estimated population standard deviation of 5). Using
the .05 level, what should the experimenter conclude? (a) Use the steps of
hypothesis testing,



This is what we have to find:

n1=
n2=
df1=
df2=
dftotal=
m1=
m2=
s^21=
s^22=
s2pooled-
s^2m1=
s^2m2=
s^2difference=
sdifference=
t=
t needed=
Decision=


Thank yo in advance for your help! I am struggling to understand this!!

n1=20
n2=30
df1=19
df2=29
m1=38
m2=35
s^21 = 3^2 =9
s^22 = 5^2 =25
s^2 pooled = 18.666667 --- see steps below
[(19)(9)+(29)(25))] / (48) = 18.666667

sdifference= sqrt(1/n1+1/n2) times sqrt(s^2) = 1.247219
s^2m1= 9/20 =0.45
s^2m2= 25/30 =0.833
s^2difference = sqrt(0.45+0.833)=1.2833
sdiidernce = 1.132843
t=(38-35)/1.327 =2.2607

t needed = (critical t with 19 deg of freedom) = 2.093
Decision = reject the null hypothesis
--------------------------------------…

H0: μ (1) = μ (2)
HA: μ (1) ≠ μ (2)

-------------------------------------
Equality of variances test:

H0: σ (1)^2 = σ (2)^2
Ha: σ (1)^2 ≠ σ (2)^2
Larger variance = 25
Smaller variance = 9

F = 2.77778
Degrees of freedom 29,19
Critical F(0.05,29,19)=2.07
Computed F > critical F, so variances are unequal. Consequently, we use the separate-variance t-test.