We would like to estimate the difference in the mean age of registered Liberals

Question

We would like to estimate the difference in the mean age of registered Liberals and registered Conservative. The mean age of a random sample of 14 registered Liberals is 43 years with a standard deviation of 7 years. The mean age of a random sample of 18 registered Conservatives is 38 years with a standard deviation of 4 years.

A) Construct a 99% confidence interval for the difference in the true mean ages of Liberals and Conservatives.

Lower limit:____________

Upper limit:____________

What is the proper interpretation of the confidence interval obtained?

B) We would like to conduct a hypothesis test at the 1% level of significance to determine whether there is a difference in the true mean age of Liberals and Conservatives. What is the rejection rule using the critical value method? Calculate the test statistic and use decide to reject H0 or fail to reject H0.

C) Could you have used the confidence interval in Question A to conduct the hypothesis test in Question B? Why or why not? If you could use the interval to conduct the test, explain what your conclusion would be and why.

Explanation / Answer

a.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=43
Standard deviation( sd1 )=7
Sample Size(n1)=14
Mean(x2)=38
Standard deviation( sd2 )=4
Sample Size(n2)=18
CI = [ ( 43-38) ±t a/2 * Sqrt( 49/14+16/18)]
= [ (5) ± t a/2 * Sqrt( 4.3889) ]
= [ (5) ± 3.012 * Sqrt( 4.3889) ]
= [-1.31 , 11.31]
b.
Given that,
mean(x)=43
standard deviation , s.d1=7
number(n1)=14
y(mean)=38
standard deviation, s.d2 =4
number(n2)=18
null, Ho: u1 = u2
alternate,there is a difference in the true mean age of liberals and conservative H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =3.012
since our test is two-tailed
reject Ho, if to < -3.012 OR if to > 3.012
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =43-38/sqrt((49/14)+(16/18))
to =2.387
| to | =2.387
critical value
the value of |t | with min (n1-1, n2-1) i.e 13 d.f is 3.012
we got |to| = 2.38667 & | t | = 3.012
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.3867 ) = 0.033
hence value of p0.01 < 0.033,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.387
critical value: -3.012 , 3.012
decision: do not reject Ho
p-value: 0.033

c.
we can conclude the decision using the confidence interval but it is a range we get
that population mean lies in the interval

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