It is desired to determine if the means of two different processes differ. Compa
ID: 3315582 • Letter: I
Question
It is desired to determine if the means of two different processes differ. Company A: n=5 x=1112.8 s-4.52 Company B:n-15 x=1115.4 s-.45 1 Perform a.05 level test assuming equal variances. . This test is best described as: A One sample t test B Paired t- test . c 2 test . D F- test -test. F 2-sample t-test Our chart number (not data value) for this problem is equal to (3 decimal places) In this problem, our data value falls into the shaded region of our properly labeled picture. What are the results of our test? A Reject Ho. The data supports Ha. B Reject Ho. The data does not support Ho. C Fail to Reject Ho. The data supports Ho. D Fail to reject Ho. The data does not support Ha.Explanation / Answer
2 sample t test
Given that,
mean(x)=1112.8
standard deviation , s.d1=4.52
number(n1)=5
y(mean)=1115.4
standard deviation, s.d2 =4.51
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.776
since our test is two-tailed
reject Ho, if to < -2.776 OR if to > 2.776
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1112.8-1115.4/sqrt((20.4304/5)+(20.3401/15))
to =-1.115
| to | =1.115
critical value
the value of |t | with min (n1-1, n2-1) i.e 4 d.f is 2.776
we got |to| = 1.11453 & | t | = 2.776
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.1145 ) = 0.327
hence value of p0.05 < 0.327,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.115
critical value: -2.776 , 2.776
decision: do not reject Ho
p-value: 0.327
option :C
fail to reject the null hypothesis,data support the Ho
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