In an article in the Journal of Advertising , Weinberger and Spotts compare the

ID: 3355882 • Letter: I

Question

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. A random sample of 400 television ads in the United Kingdom reveals that 146 use humor, while a random sample of 500 television ads in the United States reveals that 125 use humor.

Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States.

Test the hypotheses you set up in part a by using critical values and by setting equal to .10, .05, .01, and .001. How much evidence is there that the proportions of U.K. and U.S. ads using humor are different? (Round the proportion values to 3 decimal places. Round your answer to 2 decimal places.)

Set up the hypotheses needed to attempt to establish that the difference between the proportions of U.K. and U.S. ads using humor is more than .05 (five percentage points). Test these hypotheses by using a p-value and by setting a equal to .10, .05, .01, and .001. How much evidence is there that the difference between the proportions exceeds .05? (Round the proportion values to 3 decimal places. Round your z value to 2 decimal places and p-value to 4 decimal places.)

Calculate a 95 percent confidence interval for the difference between the proportion of U.K. ads using humor and the proportion of U.S. ads using humor. Interpret this interval. Can we be 95 percent confident that the proportion of U.K. ads using humor is greater than the proportion of U.S. ads using humor? (Round the proportion values to 3 decimal places. Round your answers to 4 decimal places.)

rev: 11_06_2014_QC_58708

Explanation / Answer

a.
sample one, x1 =146, n1 =400, p1= x1/n1=0.37
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
null, Ho: p1 = p2
alternate, H1: p1 != p2
b.
level of significance, = 0.1
Given that,
sample one, x1 =146, n1 =400, p1= x1/n1=0.37
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.365-0.25)/sqrt((0.301*0.7(1/400+1/500))
zo =3.74
| zo | =3.74
critical value
the value of |z | at los 0.1% is 1.645
we got |zo| =3.737 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.737 ) = 0.0002
hence value of p0.1 > 0.0002,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 3.74
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0.0002

at level of significance, = 0.05
Given that,
sample one, x1 =146, n1 =400, p1= x1/n1=0.37
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.365-0.25)/sqrt((0.301*0.7(1/400+1/500))
zo =3.74
| zo | =3.74
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =3.737 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.737 ) = 0.0002
hence value of p0.05 > 0.0002,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 3.74
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0002

At level of significance, = 0.01
Given that,
sample one, x1 =146, n1 =400, p1= x1/n1=0.37
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.365-0.25)/sqrt((0.301*0.7(1/400+1/500))
zo =3.74
| zo | =3.74
critical value
the value of |z | at los 0.01% is 2.576
we got |zo| =3.737 & | z | =2.576
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.737 ) = 0.0002
hence value of p0.01 > 0.0002,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 3.74
critical value: -2.576 , 2.576
decision: reject Ho
p-value: 0.0002

At level of significance, = 0.001
Given that,
sample one, x1 =146, n1 =400, p1= x1/n1=0.37
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.001
from standard normal table, two tailed z /2 =3.291
since our test is two-tailed
reject Ho, if zo < -3.291 OR if zo > 3.291
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.365-0.25)/sqrt((0.301*0.7(1/400+1/500))
zo =3.74
| zo | =3.74
critical value
the value of |z | at los 0.001% is 3.291
we got |zo| =3.737 & | z | =3.291
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.737 ) = 0.0002
hence value of p0.001 > 0.0002,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 3.74
critical value: -3.291 , 3.291
decision: reject Ho
p-value: 0.0002

c.
At level of significance, = 0.001
Given that,
sample one, x1 =146, n1 =400, p1= x1/n1=0.365
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, = 0.001
from standard normal table,right tailed z /2 =3.09
since our test is right-tailed
reject Ho, if zo > 3.09
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.365-0.25)/sqrt((0.301*0.699(1/400+1/500))
zo =3.737
| zo | =3.737
critical value
the value of |z | at los 0.001% is 3.09
we got |zo| =3.737 & | z | =3.09
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 3.737 ) = 0.00009
hence value of p0.001 > 0.00009,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 3.737
critical value: 3.09
decision: reject Ho
p-value: 0.00009

At level of significance, = 0.05
Given that,
sample one, x1 =146, n1 =400, p1= x1/n1=0.365
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.365-0.25)/sqrt((0.301*0.699(1/400+1/500))
zo =3.737
| zo | =3.737
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =3.737 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 3.737 ) = 0.00009
hence value of p0.05 > 0.00009,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 3.737
critical value: 1.64
decision: reject Ho
p-value: 0.00009

At level of significance, = 0.01
Given that,
sample one, x1 =146, n1 =400, p1= x1/n1=0.365
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.365-0.25)/sqrt((0.301*0.699(1/400+1/500))
zo =3.737
| zo | =3.737
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =3.737 & | z | =2.33
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 3.737 ) = 0.00009
hence value of p0.01 > 0.00009,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 3.737
critical value: 2.33
decision: reject Ho
p-value: 0.00009

At level of significance, = 0.1
Given that,
sample one, x1 =146, n1 =400, p1= x1/n1=0.365
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, = 0.1
from standard normal table,right tailed z /2 =1.28
since our test is right-tailed
reject Ho, if zo > 1.28
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.365-0.25)/sqrt((0.301*0.699(1/400+1/500))
zo =3.737
| zo | =3.737
critical value
the value of |z | at los 0.1% is 1.28
we got |zo| =3.737 & | z | =1.28
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 3.737 ) = 0.00009
hence value of p0.1 > 0.00009,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 3.737
critical value: 1.28
decision: reject Ho
p-value: 0.00009
d.
TRADITIONAL METHOD
given that,
sample one, x1 =146, n1 =400, p1= x1/n1=0.365
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.365*0.635/400) +(0.25 * 0.75/500))
=0.0309
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
margin of error = 1.645 * 0.0309
=0.0508
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.365-0.25) ±0.0508]
= [ 0.0642 , 0.1658]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =146, n1 =400, p1= x1/n1=0.365
sample two, x2 =125, n2 =500, p2= x2/n2=0.25
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.365-0.25) ± 1.645 * 0.0309]
= [ 0.0642 , 0.1658 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 0.0642 , 0.1658] contains the difference between
true population proportion P1-P2
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 difference between
true population mean P1-P2

Navigate

In an article in the Journal of Advertising , Weinberger and Spotts compare the

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.

In an article in the Journal of Advertising , Weinberger and Spotts compare the

Question

Explanation / Answer

Related Questions

Navigate