Question

Democrats and Republicans were surveyed on their support of gun control legislation using a standardized measure (high scores representing greater support of gun control). Using the data provided, calculate t. Democrats Republicans N= 50 N= 50 Mean = 6.0 Mean = 4.9 Std. Dev. = 1.2 Std. Dev. 1.4

Answer 1

Given that,
mean(x)=6
standard deviation , s.d1=1.2
number(n1)=50
y(mean)=4.9
standard deviation, s.d2 =1.4
number(n2)=50
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.677
since our test is right-tailed
reject Ho, if to > 1.677
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =6-4.9/sqrt((1.44/50)+(1.96/50))
to =4.2183
| to | =4.2183
critical value
the value of |t α| with min (n1-1, n2-1) i.e 49 d.f is 1.677
we got |to| = 4.21831 & | t α | = 1.677
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 4.2183 ) = 0.00005
hence value of p0.05 > 0.00005,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 4.2183
critical value: 1.677
decision: reject Ho
p-value: 0.00005
we have enough evidence to support the claim that Democrats and Republicans were surveyed on their support of gun control legislation
high scores representing greater support of gun control