Question

Conduct a Hypothesis test for mean comparison one-tail or two-tail (1 test on 2 main variables) on the information provided below.:

Sample 1: Mean: 29,264.10. Standard Deviation: 8,446.84. N=30.

19,400
24,380
32,500
25,400
34,395
19,500
23,570
30,930
32,000
29,000
34,000
36,800
45,000
39,500
41,830
45,000
35,800
35,950
40,000
26,810
26,995
23,710
17,999
19,000
26,000
24,000
12,875
31,000
18,000
26,579

Sample 2:

Mean: 57,640.83333. Standard Deviation: 29,995.85466. N=30

$48,950
$34,900
$102,600
$54,500
$33,900
$26,415
$39,785
$18,645
$22,995
$24,595
$30,750
$26,790
$66,500
$40,250
$89,900
$52,950
$75,150
$32,700
$113,900
$64,900
$54,600
$51,400
$113,900
$69,900
$49,700
$31,950
$54,900
$124,300
$91,100
$85,000

Unequal-variances t-test for mean comparison is the test we are told to use. Please list steps and calculations for the problem. Thank you so much!

Answer 1

Given that,
mean(x)=29264.1
standard deviation , s.d1=8446.84
number(n1)=30
y(mean)=57640.833
standard deviation, s.d2 =29995.85466
number(n2)=30
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.045
since our test is two-tailed
reject Ho, if to < -2.045 OR if to > 2.045
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =29264.1-57640.833/sqrt((71349105.9856/30)+(899751296.78384/30))
to =-4.988
| to | =4.988
critical value
the value of |t α| with min (n1-1, n2-1) i.e 29 d.f is 2.045
we got |to| = 4.98759 & | t α | = 2.045
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -4.9876 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -4.988
critical value: -2.045 , 2.045
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that difference in means between two samples.