The Municipal Transit Authority wants to know if, on weekdays, more passengers r
ID: 3390356 • Letter: T
Question
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:
n x s 8:15 a.m. train 30 323 passengers 41 8:30 a.m. train 45 passengers 45Explanation / Answer
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