Question 331543
Since analysis is on averages and the samples are considerably larger than 30, by the Central Limit theorem this makes the assumption of normally distributed averages reasonable

Ho: Mu1=Mu2
Ha: Mu1 not equal Mu2

testing at alpha=0.01

The test statistic is a t distribution since population variances are unknown, but there at two t distributions one where the pop variances are equal and one where the pop variance are unequal

You should test for equality of population variances to determine the proper t test.

Ho:  sigma1 = sigma2
Ha:  sigma1 not equal sigma2

Test statistic  {{{F=max(S1^2, S2^2)/min(S1^2,S2^2) = 89.2^2/38.7^2 = 5.312}}}
Critical value F(alpha, dof num, dof dem)=F(.01,74,84)=1.691

Since the test statistic exceeds the critical value, reject Ho and conclude that the population variances are not equal (ie cannot pool sample variances)

therefore, the test statistic for the means is 

{{{t= (Xbar1-Xbar2)/sqrt(S1^2/n1 + S2^2/n2) = (189.1-203.7)/sqrt(38.7^2/85 + 89.2^2/75) = -1.313  }}}

the degrees of freedom for this statistic is kind of messy

{{{dof = (s1^2/n1 + s2^2/n2)^2 / ((s1^2/n1)^2 /(n1-1) + (s2^2/n2)^2/(n2-1) ) =15303.77/155.79=98.2 }}} or 99 dof

pvalue = P(t<-1.313)+P(t>1.313) = 0.192   (this a two tail test)


since pvalue > alpha  ie 0.192 > 0.01  this means the a t value as extreme as  -1.313 could have come from distributions whose means are equal by pure chance ( 19.2%) 
In other words, this is possible under the null, so cannot reject Ho and conclude no evidence agains the means being equal