To assess the value of a proposed public relations campaign, the Marketing Vice
ID: 3173196 • Letter: T
Question
To assess the value of a proposed public relations campaign, the Marketing Vice President for a company asked seven consumers how much they liked the company (on a scale from 0 [do not like] to 50 [like very much]) before and after viewing the campaign's primary television advertisement. Use the following data to test whether the consumers' ratings of the company increased after viewing the television advertisement, on average: a. State the null and alternative hypotheses associated with the test. b. If alpha = 0.05, what is the critical value of the associated test statistic? c. What is the calculated value of the associated test statistic? d. State your decision about the null hypothesis by comparing the critical and calculated values of the test statistic (Parts b and c). e. Comment on the effectiveness of the primary television advertisement of the campaign.Explanation / Answer
Given that,
population mean(u)=21
sample mean, x =17
standard deviation, s =29
number (n)=38
null, H0: Ud > 0
alternate, H1: Ud < 0
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.943
since our test is left-tailed
reject Ho, if to < -1.943
we use Test Statistic
to= d/ (S/n)
Where
Value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -6
We have d = -6
Pooled variance = Calculate value of Sd= S^2 = Sqrt [ 486-(-42^2/7 ] / 6 = 6.24
to = d/ (S/n) = -2.54
Critical Value
The Value of |t | with n-1 = 6 d.f is 1.943
We got |t o| = 2.54 & |t | =1.943
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
p-value :left tail - Ha : ( p < -2.542 ) = 0.02198
hence value of p0.05 > 0.02198,here we reject Ho
ANSWERS
---------------
a.
null, H0: Ud > 0
alternate, H1: Ud < 0
b.
critical value: reject Ho, if to < -1.943
c.
test statistic: -2.54
decision: reject Ho
d.
p-value: 0.02198
e.
consumers’ ratings of the company increased after viewing the television advertisement, on average
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