Do a two-sample test for equality of means assuming unequal variances. Calculate

Question

Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors -4.25, s1 = .20, n1 = 15, x2 4.6, s2 .30, n2 15, = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick" rules for degrees of freedom.) d.f t-calculated p-value t-critical (a-2) Based on the above data choose the correct decision. Do not reject the null hypothesis OReject the null hypothesis (b-1) Comparison of average commute miles for randomly chosen students at two community colleges -21, s,-5, n1-22, x2-28, s2-7, n2-19, :05, two-tailed test (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick" rules for degrees of freedom.) d.f t-calculated p-value (b-2) Based on the above data choose the correct decision. O Reject the null hypothesis Do not reject the null hypothesis

Explanation / Answer

Given that,
mean(x)=4.25
standard deviation , s.d1=0.2
number(n1)=15
y(mean)=4.6
standard deviation, s.d2 =0.3
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, = 0.025
from standard normal table,left tailed t /2 =2.145
since our test is left-tailed
reject Ho, if to < -2.145
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =4.25-4.6/sqrt((0.04/15)+(0.09/15))
to =-3.7596
| to | =3.7596
critical value
the value of |t | with min (n1-1, n2-1) i.e 14 d.f is 2.145
we got |to| = 3.7596 & | t | = 2.145
make decision
hence value of | to | > | t | and here we reject Ho
p-value:left tail - Ha : ( p < -3.7596 ) = 0.00106
hence value of p0.025 > 0.00106,here we reject Ho
ANSWERS
---------------
a-1.

null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -3.7596
critical value: -2.145
p-value: 0.00106

a-2.

decision: reject Ho
Given that,
mean(x)=21
standard deviation , s.d1=5
number(n1)=22
y(mean)=28
standard deviation, s.d2 =7
number(n2)=19
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.101
since our test is two-tailed
reject Ho, if to < -2.101 OR if to > 2.101
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =21-28/sqrt((25/22)+(49/19))
to =-3.6316
| to | =3.6316
critical value
the value of |t | with min (n1-1, n2-1) i.e 18 d.f is 2.101
we got |to| = 3.63162 & | t | = 2.101
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.6316 ) = 0.002
hence value of p0.05 > 0.002,here we reject Ho
ANSWERS
---------------
b-1.

null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -3.6316
critical value: -2.101 , 2.101
p-value: 0.002

b-2.

decision: reject Ho
Given that,
mean(x)=147
standard deviation , s.d1=2.8
number(n1)=12
y(mean)=143
standard deviation, s.d2 =2.7
number(n2)=17
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.796
since our test is right-tailed
reject Ho, if to > 1.796
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =147-143/sqrt((7.84/12)+(7.29/17))
to =3.8452
| to | =3.8452
critical value
the value of |t | with min (n1-1, n2-1) i.e 11 d.f is 1.796
we got |to| = 3.84516 & | t | = 1.796
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 3.8452 ) = 0.00136
hence value of p0.05 > 0.00136,here we reject Ho
ANSWERS
---------------
c-1.

null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 3.8452
critical value: 1.796
p-value: 0.00136

c-2.
decision: reject Ho

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