distibvhon facility has decided to switcn to an avhmated Packaging system. The F
ID: 3362768 • Letter: D
Question
distibvhon facility has decided to switcn to an avhmated Packaging system. The Facil.wisne s h de Hermine, , at tne.. .1wvel, If the mean packag ing hmes, given in seconds per pactage, are dieevent metmok that tha system opfers, Mernod A and tor the hvo packaging Hetnod Method p 22.49 22 1S 239 21.41 20.52 20.8 28 07 19.02 24.32 22. 8 25. OS 24.08 9.19 22.3 27.83 23.ai 2 0.18 25.24 20 01 lo. 3 23.3 20.31 Assume that the packaging hmes are indepandent and normally disthibutedl, but that the poplation vartances are unknon l state tha null and alter nahve hl pomere, indicate whe a one-taled or hwo-tailad tes+ 2. compute any ecessary ummar stahshes (uch as sampte wans ample vanan cec, and the pooed sampe vaiance ) fr the dataExplanation / Answer
Given that,
mean(x)=21.2414
standard deviation , s.d1=2.1689
number(n1)=14
y(mean)=24.0385
standard deviation, s.d2 =2.7507
number(n2)=13
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.1
from standard normal table, two tailed t /2 =1.782
since our test is two-tailed
reject Ho, if to < -1.782 OR if to > 1.782
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =21.2414-24.0385/sqrt((4.70413/14)+(7.56635/13))
to =-2.9193
| to | =2.9193
critical value
the value of |t | with min (n1-1, n2-1) i.e 12 d.f is 1.782
we got |to| = 2.9193 & | t | = 1.782
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.9193 ) = 0.013
hence value of p0.1 > 0.013,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.9193
critical value: -1.782 , 1.782
decision: reject Ho
p-value: 0.013
the mean packaging method that the system offers is diffrent between methods
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