Question

The U.S. Census Bureau collects data on the ages of married people. Suppose that eight married couples are randomly selected and have the ages given in the following table. Determine the 95% confidence interval for the true mean difference between the ages of married males and married females.

Let d=(age of husband)−(age of wife). Assume that the ages are normally distributed for the populations of both husbands and wives in the U.S.

Husband	64	40	46	35	37	67	46	61
Wife	71	33	58	37	43	72	55	55

Step 1 of 4:

Find the mean of the paired differences,  d‾. Round your answer to one decimal place.

Step 2 of 4:

Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 3 of 4:

Find the standard deviation of the paired differences to be used in constructing the confidence interval. Round your answer to one decimal place.

Step 4 of 4:

Construct the 95% confidence interval. Round your answers to one decimal place.

Answer 1

a.
Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.365
since our test is two-tailed
reject Ho, if to < -2.365 OR if to > 2.365
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -3.5
We have d = -3.5
pooled variance = calculate value of Sd= √S^2 = sqrt [ 424-(-28^2/8 ] / 7 = 6.824
to = d/ (S/√n) = -1.451
critical Value
the value of |t α| with n-1 = 7 d.f is 2.365
we got |t o| = 1.451 & |t α| =2.365
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 != -1.4506 ) = 0.1902
hence value of p0.05 < 0.1902,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: -1.451
critical value: reject Ho, if to < -2.365 OR if to > 2.365
decision: Do not Reject Ho
p-value: 0.1902
we do not have enough evidence to support the claim that the true mean difference between the ages of married males and married females.
b.
Confidence Interval
CI = d ± t a/2 * (Sd/ Sqrt(n))
Where,
d = ∑ di/n
Sd = Sqrt( ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
d = ( ∑ di/n ) =-28/8=-3.5
Pooled Sd( Sd )= Sqrt [ 424- (-28^2/8 ] / 7 = 6.824
Confidence Interval = [ -3.5 ± t a/2 ( 3.94/ Sqrt ( 8) ) ]
= [ -3.5 - 2.365 * (2.413) , -3.5 + 2.365 * (2.413) ]
= [ -9.206 , 2.206 ]
Answers:
mean of paired difference = -3.5
critical value is to < -2.365 OR if to > 2.365
standard deviation = 6.824
confidence interval is [ -9.206 , 2.206 ]