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: x⎯⎯1 = 4, s1 = .20, n1 = 15, x⎯⎯2 = 4.25, s2 = .30, n2 = 15, α = .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. 24 24 Correct t-calculated 3.2225 3.2225 Incorrect p-value .0018 .0018 Incorrect t-critical 2.064 2.064 Incorrect (a-2) Based on the above data choose the correct decision. Reject the null hypothesis Do not reject the null hypothesis (b-1) Comparison of average commute miles for randomly chosen students at two community colleges: x⎯⎯1 = 17, s1 = 5, n1 = 22, x⎯⎯2 = 21, s2 = 7, n2 = 19, α = .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. 32 32 Correct t-calculated 3.1128 3.1128 Incorrect p-value .0039 .0039 Incorrect t-critical +/- 2.037 2.037 Correct (b-2) Based on the above data choose the correct decision. Do not reject the null hypothesis Reject the null hypothesis (c-1) Comparison of credits at time of graduation for randomly chosen accounting and economics students: x⎯⎯1 = 141, s1 = 2.8, n1 = 12, x⎯⎯2 = 138, s2 = 2.7, n2 = 17, α = .05, right-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. 23 23 Correct t-calculated .9613 .9613 Incorrect p-value .1732 .1732 Incorrect t-critical 1.714 1.714 Correct (c-2) Based on the above data choose the correct decision. Reject the null hypothesis Do not reject the null hypothesis

Answer 1

a.
Given that,
mean(x)=4
standard deviation , s.d1=0.2
number(n1)=15
y(mean)=4.25
standard deviation, s.d2 =0.3
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.025
from standard normal table,left tailed t α/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 =4-4.25/sqrt((0.04/15)+(0.09/15))
to =-2.685
| to | =2.685
critical value
the value of |t α| with min (n1-1, n2-1) i.e 14 d.f is 2.145
we got |to| = 2.68543 & | t α | = 2.145
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:left tail - Ha : ( p < -2.6854 ) = 0.00888
hence value of p0.025 > 0.00888,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -2.685
critical value: -2.145
decision: reject Ho
p-value: 0.00888
we have enough evidence to support the claim that mean of sample 1 is less than mean of sample 2.
b.
Given that,
mean(x)=17
standard deviation , s.d1=5
number(n1)=22
y(mean)=21
standard deviation, s.d2 =7
number(n2)=19
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/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 =17-21/sqrt((25/22)+(49/19))
to =-2.075
| to | =2.075
critical value
the value of |t α| with min (n1-1, n2-1) i.e 18 d.f is 2.101
we got |to| = 2.07521 & | t α | = 2.101
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 != -2.0752 ) = 0.053
hence value of p0.05 < 0.053,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.075
critical value: -2.101 , 2.101
decision: do not reject Ho
p-value: 0.053
we do not have enough evidence to support the claim that difference of means between two samples.
c.
Given that,
mean(x)=141
standard deviation , s.d1=2.8
number(n1)=12
y(mean)=138
standard deviation, s.d2 =2.7
number(n2)=17
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/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 =141-138/sqrt((7.84/12)+(7.29/17))
to =2.884
| to | =2.884
critical value
the value of |t α| with min (n1-1, n2-1) i.e 11 d.f is 1.796
we got |to| = 2.88387 & | t α | = 1.796
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 2.8839 ) = 0.00743
hence value of p0.05 > 0.00743,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2.884
critical value: 1.796
decision: reject Ho
p-value: 0.00743
we have enough evidence to support the claim that mean of sample 1 is greater than mean of sample 2.