Question

n1=37, n2=37, s1= 7.4345, s2= 3.4927, x1= 23.237, x2 = 22.526 Please perform: One Hypothesis test, an F test for the equality of the variances of travel Times and the second test is a T-test for the equality of the means of travel times in MINUTES. The F test must be performed first in order to select either Case1 or Case 2 for the T-test. Then perform the Required T-test (either case 1 or 2 depending on your findings of the F-test). use p value as rejection rule for both tests!!, and use the 5 steps please please help

-the pvalue has to be used as rejection rule, and an f test then a t test (case 1 or 2 depending on f test results). The answers have to be in an interval. Please help me. I need this to review and study. I dont know how to get them in an interval. please make sure the p value is used

*FOR BOTH TESTS THE P VALUE NEEDS TO BE IN A RANGE AND THEN IF REJECTED OR NOT. I KNOW THIS HAS BEEN POSTED BEFORE BUT no one has helped me with the correct set up yet, aka the range, also a skewed plot for f test. please provide a range for the p value. i really need help

Answer 1

a.
Given that,
mean(x)=23.237
standard deviation , s.d1=7.4345
number(n1)=37
y(mean)=22.526
standard deviation, s.d2 =3.4927
number(n2)=37
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.028
since our test is two-tailed
reject Ho, if to < -2.028 OR if to > 2.028
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =23.237-22.526/sqrt((55.27179/37)+(12.19895/37))
to =0.5265
| to | =0.5265
critical value
the value of |t α| with min (n1-1, n2-1) i.e 36 d.f is 2.028
we got |to| = 0.52652 & | t α | = 2.028
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.5265 ) = 0.602
hence value of p0.05 < 0.602,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.5265
critical value: -2.028 , 2.028
decision: do not reject Ho
p-value: 0.602
we do not have enough evidence to support the claim that difference of means.
b.
Given that,
sample 1
s1^2=55.271, n1 =37
sample 2
s2^2 =12.198, n2 =37
null, Ho: σ^2 = σ^2
alternate, H1: σ^2 != σ^2
level of significance, α = 0.05
from standard normal table, two tailed f α/2 =1.942
since our test is two-tailed
reject Ho, if F o < -1.942 OR if F o > 1.942
we use test statistic fo = s1^1/ s2^2 =55.271/12.198 = 4.53
| fo | =4.53
critical value
the value of |f α| at los 0.05 with d.f f(n1-1,n2-1)=f(36,36) is 1.942
we got |fo| =4.531 & | f α | =1.942
make decision
hence value of | fo | > | f α| and here we reject Ho
ANSWERS
---------------
null, Ho: σ^2 = σ^2
alternate, H1: σ^2 != σ^2
test statistic: 4.53
critical value: -1.942 , 1.942
decision: reject Ho
p value is 0.0000089
we have enough evidence to support the claim that difference of variances.