The Municipal Transit Authority wants to know if on weekdays

The Municipal Transit Authority wants to know if, on weekdays, more passengers ride the northbound blue line train towards the city center that departs at 8:30 a.m. than the one that departs at 8:15 a.m. The following sample statistics are assembled by the Transit Authority.

(a) First, construct the 90% confidence interval for the difference in the mean number of daily travelers on the 8:15 train and the mean number of daily travelers on the 8:30 train:

(b) Next, test at the 5% level of significance whether the data provide sufficient evidence to conclude that more passengers ride the 8:30 train:

(c) State null and alternative hypotheses:

(d) Determine distribution of test statistic and compute its value:

(e) Construct the rejection region:

(f) Make your decision:

(g) State your conclusion:

(h) Compute the p-value (observed level of significance) for this test:

Solution

a.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=323
Standard deviation( sd1 )=41
Sample Size(n1)=30
Mean(x2)=356
Standard deviation( sd2 )=45
Sample Size(n1)=45
CI = [ ( 323-356) ±t a/2 * Sqrt( 1681/30+2025/45)]
= [ (-33) ± t a/2 * Sqrt( 101.03) ]
= [ (-33) ± 1.699 * Sqrt( 101.03) ]
= [-50.08 , -15.92]

b.
Set Up Hypothesis
Null, There Is No-Significance between them Ho: u1 > u2
Alternate, more passengers ride the 8:30 train - H1: u1 < u2
Test Statistic
X(Mean)=323
Standard Deviation(s.d1)=41 ; Number(n1)=30
Y(Mean)=356
Standard Deviation(s.d2)=45; Number(n2)=45
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =323-356/Sqrt((1681/30)+(2025/45))
to =-3.28
| to | =3.28
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 29 d.f is , Reject if to< - 1.699
We got |to| = 3.28308 & | t | = 1.699
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value:Left Tail - Ha : ( P < -3.2831 ) = 0.00134
Hence Value of P0.05 > 0.00134,Here we Reject Ho

we have evidence that more passengers ride the 8:30 train