a In order to test whether there is a difference in mean mai

a) In order to test whether there is a difference in mean maintenance costs between buses with Diesel engines and those with gasoline engines, use Excel to carry out a “t-Test: Two-Sample Assuming Equal Variance”.

b) Based on the t-test, what is your decision regarding the null hypothesis (“reject” or “do not reject”)? Indicate on which part of the Excel output you base your decision.

c) Explain your answer to part (b) at a level that can be understood by a high school graduate who may or may not have taken some college-level coursework, but has not taken a college-level course in statistics.

Solution

Set Up Hypothesis
Null Hypothesis, There Is NoSignificance between them Ho: u1 = u2
Alternative Hypothesis, There Is Significance between themH1: u1 != u2
Test Statistic
X (Mean)=449.926; Standard Deviation (s.d1)=69.2059
Number(n1)=27
Y(Mean)=454.263; Standard Deviation(s.d2)=60.564
Number(n2)=19
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (26*4789.4566 + 18*3667.9981) / (46- 2 )
S^2 = 4330.6781
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=449.926-454.263/Sqrt((4330.6781( 1 /27+ 1/19 ))
to=-4.337/19.706
to=-0.2201
| to | =0.2201
Critical Value
The Value of |t | with (n1+n2-2) i.e 44 d.f is 2.015
We got |to| = 0.2201 & | t | = 2.015
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -0.2201 ) = 0.8268
Hence Value of P0.05 < 0.8268,Here We Do not Reject Ho

There is no diffrence in mean maintenance costs between buses with Diesel engines and those with gasoline engines