Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

at the time she was hired at the time she was hired at the time she was hired at

ID: 3355823 • Letter: A

Question

at the time she was hired at the time she was hired at the time she was hired at the time she was hired at the time she was hired at the time she was hired at the time she was hired at the time she was hired at the time she was hired Excel File Edit View Insert Format Tools Data Window Help newconnect.mheducation.com Exam 3_Chapters 8-11 Saved Help Save & Exit Submit The null and alternate hypotheses are: 6 A random sample of 27 items from the first population showed a mean of 114 and a standard deviation of 15. A sample of 15 items for the second population showed a mean of 106 and a standard deviation of 9. Use tht 0.03 significant level 00:35:23 a. Find the degrees of freedom for unequal variance test. (Round down your answer to the nearest whole number.) rees of freedom S0 b. State the decision rule for O.025 significance level. (Round your answer to 3 decimal places) 56 57 Reject HO if t> 59 61 62 63 64 65 c. Compute the value of the test statistic. (Round your answer to 3 decimal places.) 67 68 69 70 Value of the test statistic

Explanation / Answer

Given that,
mean(x)=114
standard deviation , s.d1=15
number(n1)=27
y(mean)=106
standard deviation, s.d2 =9
number(n2)=15
null, Ho: u1 <= u2
alternate, H1: u1 > u2
level of significance, = 0.03
from standard normal table,right tailed t /2 =2.046
since our test is right-tailed
reject Ho, if to > 2.046
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =114-106/sqrt((225/27)+(81/15))
to =2.159
| to | =2.159
critical value
the value of |t | with min (n1-1, n2-1) i.e 14 d.f is 2.046
we got |to| = 2.15875 & | t | = 2.046
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 2.1587 ) = 0.02436
hence value of p0.03 > 0.02436,here we reject Ho
ANSWERS
---------------
a. with min (n1-1, n2-1) i.e 14
b.

for alpha = 0.03, reject Ho, if to > 2.046

level of significance, = 0.025
from standard normal table,right tailed t /2 =2.145
since our test is right-tailed
reject Ho, if to > 2.145

c.
test statistic: 2.159