A psychologist obtained recordings of election-night acceptance speeches of seve

Question

A psychologist obtained recordings of election-night acceptance speeches of seven newly elected representatives to the U.S. Congress and counted the number of minutes devoted to urban problems in these speeches. The four of these representatives from rural districts devoted 5, 0, 3, and 4 minutes to urban problems, and the three representatives from urban districts devoted 11, 11, and 14 minutes to urban problems. Using the.05 significance level, did the amount of time devoted to urban problems in the acceptance speech differ according to the type of district the representative represented? Use the five steps of hypothesis testing and the t-test for independent means. Calculate Effect Size for Extra Credit Number of Minutes in Speeches 0 14 4

Explanation / Answer

Question1.
Given that,
mean(x)=3
standard deviation , s.d1=2.1602
number(n1)=4
y(mean)=12
standard deviation, s.d2 =1.7321
number(n2)=3
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =4.3
since our test is two-tailed
reject Ho, if to < -4.3 OR if to > 4.3
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =3-12/sqrt((4.66646/4)+(3.00017/3))
to =-6.114
| to | =6.114
critical value
the value of |t | with min (n1-1, n2-1) i.e 2 d.f is 4.3
we got |to| = 6.11429 & | t | = 4.3
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -6.1143 ) = 0.026
hence value of p0.05 > 0.026,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -6.114
critical value: -4.3 , 4.3
decision: reject Ho
p-value: 0.026

Question2.
Given that,
mean(x)=13
standard deviation , s.d1=1.4142
number(n1)=4
y(mean)=9
standard deviation, s.d2 =2.1602
number(n2)=4
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =2.35
since our test is right-tailed
reject Ho, if to > 2.35
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =13-9/sqrt((1.99996/4)+(4.66646/4))
to =3.098
| to | =3.098
critical value
the value of |t | with min (n1-1, n2-1) i.e 3 d.f is 2.35
we got |to| = 3.09844 & | t | = 2.35
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 3.0984 ) = 0.02668
hence value of p0.05 > 0.02668,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 3.098
critical value: 2.35
decision: reject Ho
p-value: 0.02668

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