Question 332017
Let Mu(n)= Population mean rating of new trainer
Let Mu(e)= Population mean rating of existing trainer
Let Xbar(n) = sample average rating for new trainer
Let Xbar(e) = sample average rating for existing trainer
Let S(n)=sample std dev for new trainer
Let S(e)=sample std dev for existing trainer
Let n(n)= sample size for new trainer
Let n(e)= sample size for exiting trainer
--
to answer
Did the new leader of the Tuesday night class receive a lower approval rating than the leader of the Thursday night class? 
--
Ho: Mu(n)=Mu(e)
Ha: Mu(n) < Mu(e)

at alpha =0.05
lower tail test
---
Find the test statistic
This is a two independent samples t test.  There are two t tests, one if the population variances are equal and one if the population variances are not equal
----
To determine the t test to use, you should conduct a side hypothesis test on the population variances
Ho: Variance (n) =Variance (e)
Ha: Variance (n) not equla variance (e)

test statistic:  F=max(var(n),var(e))/min(var(n),var(e)) = (7.4/6.3)^2  =1.38
critical value is F(alpha, dof numerator, dof denominator) = F(.05,37,40)=1.71

since test statistic=1.38 < critical value=1.71 cannot reject Ho: variances are equal.
---
Since we now established that its likely the pop variances are equal, we may pool the asmple variances

pooled sample variance = {{{Sp^2=((38-1)*7.4^2+(41-1)*6.3^3))}}}/(38+41-2)=46.93
----
test statistic: {{{ t=(Xbar(n)-Xbar(e))/(Sp*sqrt(1/n(n) + 1/n(e)))}}}
{{{t=(87.3-89.7)/(6.85*sqrt(1/38+1/41))}}} =-2.4/1.54=-1.558

critical value = t(1-0.05,38+41-2)=-1.66

Since our test statistic= -1.558 > critical value = -1.66, it falls in the fail to reject Ho.

pvalue = P(t<=-1.558) = 0.062 > alpha =0.05
---
Conclusion

Cannot reject the hypothesis that the mean ratings between trainers is the same.