A popular theory is that presidential candidates have an advantage if they are t

Question

A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are heights (in centimeters) of randomly selected presidents along with the heights of their main opponents. Complete parts (a) and (b) below. Height left parenthesis cm right parenthesis of PresidentHeight (cm) of President 193 187 184 177 200 175 Height left parenthesis cm right parenthesis of Main OpponentHeight (cm) of Main Opponent 181 185 174 174 196 181 Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)? What is the confidence interval?

Explanation / Answer

a.
Given that,
mean(x)=186
standard deviation , s.d1=9.5079
number(n1)=6
y(mean)=181.833
standard deviation, s.d2 =8.1833
number(n2)=6
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =2.02
since our test is right-tailed
reject Ho, if to > 2.02
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =186-181.833/sqrt((90.40016/6)+(66.9664/6))
to =0.814
| to | =0.814
critical value
the value of |t | with min (n1-1, n2-1) i.e 5 d.f is 2.02
we got |to| = 0.81366 & | t | = 2.02
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 0.8137 ) = 0.22642
hence value of p0.05 < 0.22642,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 0.814
critical value: 2.02
decision: do not reject Ho
p-value: 0.22642
we do not have enough evidence to support the claim that they are taller than their main opponents
b.
TRADITIONAL METHOD
given that,
mean(x)=186
standard deviation , s.d1=9.5079
number(n1)=6
y(mean)=181.833
standard deviation, s.d2 =8.1833
number(n2)=6
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((90.4/6)+(66.97/6))
= 5.12
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table,right tailedand
value of |t | with min (n1-1, n2-1) i.e 5 d.f is 2.02
margin of error = 2.015 * 5.12
= 10.32
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (186-181.833) ± 10.32 ]
= [-6.15 , 14.49]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=186
standard deviation , s.d1=9.5079
sample size, n1=6
y(mean)=181.833
standard deviation, s.d2 =8.1833
sample size,n2 =6
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 186-181.833) ± t a/2 * sqrt((90.4/6)+(66.97/6)]
= [ (4.17) ± t a/2 * 5.12]
= [-6.15 , 14.49]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-6.15 , 14.49] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

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