According to literature on brand loyalty, consumers who are loyal to a brand are

Question

According to literature on brand loyalty, consumers who are loyal to a brand are likely to consistently select the same product. This type of consistency could come from a positive childhood association. To examine brand loyalty among fans of the Chicago Cubs, 372 Cubs fans among patrons of a restaurant located in Wrigleyville were surveyed prior to a game at Wrigley Field, the Cubs' home field. The respondents were classified as "die-hard fans" or "less loyal fans." Of the 133 die-hard fans, 91.0% reported that they had watched or listened to Cubs games when they were children. Among the 239 less loyal fans, 67.4% said that they watched or listened as children. (Let D = pdie-hard pless loyal.)

(a) Find the numbers of die-hard Cubs fans who watched or listened to games when they were children. Do the same for the less loyal fans. (Round your answers to the nearest whole number.) die-hard fans less loyal fans

(b) Use a one sided significance test to compare the die-hard fans with the less loyal fans with respect to their childhood experiences relative to the team. (Use your rounded values from part (a). Use = 0.01. Round your z-value to two decimal places and your P-value to four decimal places.) z = P-value = Conclusion Reject the null hypothesis, there is not significant evidence that a higher proportion of die hard Cubs fans watched or listened to Cubs games as children. Fail to reject the null hypothesis, there is significant evidence that a higher proportion of die hard Cubs fans watched or listened to Cubs games as children. Fail to reject the null hypothesis, there is not significant evidence that a higher proportion of die hard Cubs fans watched or listened to Cubs games as children. Reject the null hypothesis, there is significant evidence that a higher proportion of die hard Cubs fans watched or listened to Cubs games as children.

(c) Express the results with a 95% confidence interval for the difference in proportions. (Round your answers to three decimal places.) ,

Explanation / Answer

a.
toatl =372 divided into die hard fans and less loyal fans
133 die-hard fans, 91.0% reported that they had watched or listened to Cubs games when they were children
the 239 less loyal fans, 67.4% said that they watched or listened as children
b.
Given that,
sample one, x1 =133, n1 =372, p1= x1/n1=0.358
sample two, x2 =239, n2 =372, p2= x2/n2=0.642
null, Ho: p1 = p2
alternate, H1: p1 < p2
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.358-0.642)/sqrt((0.5*0.5(1/372+1/372))
zo =-7.772
| zo | =7.772
critical value
the value of |z | at los 0.01% is 2.326
we got |zo| =7.772 & | z | =2.326
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -7.7723 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 < p2
test statistic: -7.772
critical value: -2.326
decision: reject Ho
p-value: 0

Reject the null hypothesis, there is not significant evidence that a higher proportion of die hard Cubs fans watched or listened to Cubs games as children.
c.
TRADITIONAL METHOD
given that,
sample one, x1 =133, n1 =372, p1= x1/n1=0.358
sample two, x2 =239, n2 =372, p2= x2/n2=0.642
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*0.642/372) +(0.642 * 0.358/372))
=0.035
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.035
=0.069
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.358-0.642) ±0.069]
= [ -0.354 , -0.216]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =133, n1 =372, p1= x1/n1=0.358
sample two, x2 =239, n2 =372, p2= x2/n2=0.642
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 = [ (0.358-0.642) ± 1.96 * 0.035]
= [ -0.354 , -0.216 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ -0.354 , -0.216] 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

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