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 143 use humor, while a random sample of 500 television ads in the United States reveals that 126 use humor.

(a) 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.

(b) Test the hypotheses you set up in part a by using critical values and by setting α 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.)

(c) 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 α 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.) (d) 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.)

Answer 1

Given that,
sample one, x1 =143, n1 =400, p1= x1/n1=0.358
sample two, x2 =126, n2 =500, p2= x2/n2=0.252
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.1
from standard normal table, two tailed z α/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 =(0.358-0.252)/sqrt((0.299*0.701(1/400+1/500))
zo =3.436
| zo | =3.436
critical value
the value of |z α| at los 0.1% is 1.645
we got |zo| =3.436 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.4356 ) = 0.0006
hence value of p0.1 > 0.0006,here we reject Ho
ANSWERS
---------------
a.
null, Ho: p1 = p2
alternate, H1: p1 != p2
b.
i.
level of significance =0.10
test statistic: 3.436
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0.0006
we have enough evidence to support the claim that whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States.
ii.
level of significance =0.05
Given that,
sample one, x1 =143, n1 =400, p1= x1/n1=0.358
sample two, x2 =126, n2 =500, p2= x2/n2=0.252
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/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 =(0.358-0.252)/sqrt((0.299*0.701(1/400+1/500))
zo =3.436
| zo | =3.436
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =3.436 & | z α | =1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.4356 ) = 0.0006
hence value of p0.05 > 0.0006,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 3.436
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0006
iii.
level of significance =0.01
Given that,
sample one, x1 =143, n1 =400, p1= x1/n1=0.358
sample two, x2 =126, n2 =500, p2= x2/n2=0.252
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.01
from standard normal table, two tailed z α/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 =(0.358-0.252)/sqrt((0.299*0.701(1/400+1/500))
zo =3.436
| zo | =3.436
critical value
the value of |z α| at los 0.01% is 2.576
we got |zo| =3.436 & | z α | =2.576
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.4356 ) = 0.0006
hence value of p0.01 > 0.0006,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 3.436
critical value: -2.576 , 2.576
decision: reject Ho
p-value: 0.0006
iv.
level of significance =0.001
Given that,
sample one, x1 =143, n1 =400, p1= x1/n1=0.358
sample two, x2 =126, n2 =500, p2= x2/n2=0.252
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.001
from standard normal table, two tailed z α/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 =(0.358-0.252)/sqrt((0.299*0.701(1/400+1/500))
zo =3.436
| zo | =3.436
critical value
the value of |z α| at los 0.001% is 3.291
we got |zo| =3.436 & | z α | =3.291
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.4356 ) = 0.0006
hence value of p0.001 > 0.0006,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 3.436
critical value: -3.291 , 3.291
decision: reject Ho
p-value: 0.0006
c.
p value is 0.0006 at setting α equal to .10, .05, .01, and .001
d.
TRADITIONAL METHOD
given that,
sample one, x1 =143, n1 =400, p1= x1/n1=0.358
sample two, x2 =126, n2 =500, p2= x2/n2=0.252
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( (0.358*0.643/400) +(0.252 * 0.748/500))
=0.031
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.031
=0.06
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.358-0.252) ±0.06]
= [ 0.045 , 0.166]
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DIRECT METHOD
given that,
sample one, x1 =143, n1 =400, p1= x1/n1=0.358
sample two, x2 =126, n2 =500, p2= x2/n2=0.252
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 = [ (0.358-0.252) ± 1.96 * 0.031]
= [ 0.045 , 0.166 ]
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interpretations:
1) we are 95% sure that the interval [ 0.045 , 0.166] 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