Corporate advertising tries to enhance the image of the corporation. A study com

Question

Corporate advertising tries to enhance the image of the corporation. A study compared two ads from two sources, the Wall Street Journal and the National Enquirer. Subjects were asked to pretend that their company was considering a major investment in Performax, the fictitious sportswear firm in the ads. Each subject was asked to respond to the question "How trustworthy was the source in the sportswear company ad for Performax?" on a 7-point scale. Higher values indicated more trustworthiness. Here is a summary of the results.

Find the two-sample pooled t statistic. Then formulate the problem as an ANOVA and report the results of this analysis. Verify that F = t 2.

Explanation / Answer

PART 1

Given that,
mean(x)=4.77
standard deviation , s.d1=1.5
number(n1)=66
y(mean)=2.43
standard deviation, s.d2 =1.64
number(n2)=61
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =1.979
since our test is two-tailed
reject Ho, if to < -1.979 OR if to > 1.979
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (65*2.25 + 60*2.6896) / (127- 2 )
s^2 = 2.461
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=4.77-2.43/sqrt((2.461( 1 /66+ 1/61 ))
to=2.34/0.2786
to=8.3984
| to | =8.3984
critical value
the value of |t | with (n1+n2-2) i.e 125 d.f is 1.979
we got |to| = 8.3984 & | t | = 1.979
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 8.3984 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 8.3984
critical value: -1.979 , 1.979
decision: reject Ho
p-value: 0

PART 2

Given that,
sample 1
s1^2=1.6896, n1 =61
sample 2
s2^2 =2.25, n2 =66
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
level of significance, = 0.01
from standard normal table, two tailed f /2 =1.929
since our test is two-tailed
reject Ho, if F o < -1.929 OR if F o > 1.929
we use test statistic fo = s1^1/ s2^2 =1.6896/2.25 = 0.75
| fo | =0.75
critical value
the value of |f | at los 0.01 with d.f f(n1-1,n2-1)=f(60,65) is 1.929
we got |fo| =0.751 & | f | =1.929
make decision
hence value of |fo | < | f | and here we do not reject Ho
ANSWERS
---------------
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
test statistic: 0.75
critical value: -1.929 , 1.929
decision: do not reject Ho

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