2B of 40 (27 complete) This Test: 40 pts and 12 expert inspectors to determine w

Question

2B of 40 (27 complete) This Test: 40 pts and 12 expert inspectors to determine whether there is a difference in the variability of detecting defective products. The Excel resuts for the data are provided below F-Test Two Sample for Variances 32 83383 20 583333 746969697 3299242424 . 12 2 264064294 PFs one-tail 2.81793047 A. even though the lower tal rejection region is unknown, the decision rule for the upper-tail the nejection negion is i#F> 2 82 them eject HO B. even though the lower tail rejection region is unknown, the decision nule for the upper-talil the nejection region is if F

Explanation / Answer

Given that,
sample 1
s1^2=74.696, n1 =12
sample 2
s2^2 =32.9924, n2 =12
null, Ho: ^2 =^2
alternate, H1: ^2 > ^2
level of significance, = 0.05
from standard normal table,right tailed f /2 =2.818
since our test is right-tailed
we use test statistic fo = s1^1/ s2^2 =74.696/32.9924 = 2.26
| fo | =2.26
critical value
the value of |f | at los 0.05 with d.f f(n1-1,n2-1)=f(11,11) is 2.818
we got |fo| =2.264 & | f | =2.818
make decision
hence value of |fo | < | f | and here we do not reject Ho
ANSWERS
---------------
null, Ho: ^2 = ^2
alternate, H1: ^2 > ^2
test statistic: 2.26
critical value: 2.818
decision: do not reject Ho
Even though the lower tail rejection region is unknown,the decision rule for the upper tail rejection region is F <2.818 reject ho

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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