The average mpg usage for a 2009 Toyota Prius for a sample of 18 tanks of gas wa

Question

The average mpg usage for a 2009 Toyota Prius for a sample of 18 tanks of gas was 44, deviation of 1.8. For a 2009 Honda Insight, the average mpg usage for a sample of 18 43.1 with a standard deviation of 2.7. Assuming equal variances, at alpha = 0.01, is the true mean mpg lower for the Honda Choose the appropriate hypotheses. a. H_0: mu_toy - mu_hon lessthanorequalto 0 vs. H_1: mu_toy - mu_hon > 0. Reject H_0 if t_calc > 2.441 b. H_0: mu_toy - mu_hon lessthanorequalto 0 vs. H_1: mu_toy - mu_hon > 0. Reject H_0 if t_calc

Explanation / Answer

Given that,
mean(x)=44
standard deviation , s.d1=1.8
number(n1)=18
y(mean)=43.1
standard deviation, s.d2 =2.7
number(n2)=18
null, Ho: u1 < u2
alternate, H1: u1 > u2
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.441
since our test is right-tailed
reject Ho, if to > 2.441
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (17*3.24 + 17*7.29) / (36- 2 )
s^2 = 5.265
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=44-43.1/sqrt((5.265( 1 /18+ 1/18 ))
to=0.9/0.7649
to=1.1767
| to | =1.1767
critical value
the value of |t | with (n1+n2-2) i.e 34 d.f is 2.441
we got |to| = 1.1767 & | t | = 2.441
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: right tail -ha : ( p > 1.1767 ) = 0.12374
hence value of p0.01 < 0.12374,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 < u2
alternate, H1: u1 > u2
test statistic: 1.1767
critical value: 2.441
decision: do not reject Ho
p-value: 0.12374

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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