b) What is the p-value? (3) 2) A school district is trying to decrease the uncer

Question

b) What is the p-value? (3) 2) A school district is trying to decrease the uncertainty concerning pick-up times on a particular school bus route. Due to uncertainty in traffic patterns, the school district examines two alternative routes. In a sample of 41 days on alternative route #1, the average driving time for the school bus is min e with a sta drive isminutes.In a sample of61 days using alternative route #2, the average s with a standard deviation of 11 minutes. ce tetto see Talternativeroute #2 has alower vanancethan At the 1% level of alternative route #1. (20) a) b) Is the p-value greater or less than 5%? How do you know? (10)

Explanation / Answer

PART A. TEST FOR SIGNIFICANCE OF VARIANCE
Given that,
sample 1
s1^2=15^2 = 225, n1 =41
sample 2
s2^2 =11^2 = 121, n2 =61
null, Ho: sigma^2 = sigma^2
alternate, route#2 has lower variance than route #1 H1: sigma^2 > sigma^2
level of significance, alpha = 0.01
from standard normal table,right tailed f alpha/2 =1.936
since our test is right-tailed
we use test statistic fo = s1^1/ s2^2 =225/121 = 1.86
| fo | =1.86
critical value
the value of |f alpha| at los 0.01 with d.f f(n1-1,n2-1)=f(40,60) is 1.936
we got |fo| =1.86 & | f alpha | =1.936
make decision
hence value of |fo | < | f alpha | and here we do not reject Ho
ANSWERS
---------------
null, Ho: sigma^2 = sigma^2
alternate,route#2 has lower variance than route #1 H1: sigma^2 > sigma^2
test statistic: 1.86
critical value: 1.936
decision: do not reject Ho

PART B.

p-value : 0.014

we don't have evidence that route#2 has lower variance than route #1

PART C.

TEST FOR SIGNIFICANCE OF MEAN

Given that,
mean(x)=40
standard deviation , s.d1=15
number(n1)=41
y(mean)=45
standard deviation, s.d2 =11
number(n2)=61
null, Ho: u1 = u2
alternate, driving time of route#1 is less than driving time of alternative route #2 H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.684
since our test is left-tailed
reject Ho, if to < -1.684
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =40-45/sqrt((225/41)+(121/61))
to =-1.8292
| to | =1.8292
critical value
the value of |t | with min (n1-1, n2-1) i.e 40 d.f is 1.684
we got |to| = 1.82923 & | t | = 1.684
make decision
hence value of | to | > | t | and here we reject Ho
p-value:left tail - Ha : ( p < -1.8292 ) = 0.03741
hence value of p0.05 > 0.03741,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -1.8292
critical value: -1.684
decision: reject Ho
we have evidence that driving time of route#1 is less than driving time of alternative route #2

PART D. p-value: 0.03741

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.