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, 375 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 131 die-hard fans, 94.7% reported that they had watched or listened to Cubs games when they were children. Among the 244 less loyal fans, 66.8% 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.)

(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.)

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 =375 divided into die hard fans and less loyal fans
131 die-hard fans, 94.7%% reported that they had watched or listened to Cubs games when they were children
the 244 less loyal fans, 66.8% said that they watched or listened as children
b.
Given that,
sample one, x1 =131, n1 =375, p1= x1/n1=0.349
sample two, x2 =244, n2 =375, p2= x2/n2=0.651
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 =(0.349-0.651)/sqrt((0.5*0.5(1/375+1/375))
zo =-8.252
| zo | =8.252
critical value
the value of |z | at los 0.01% is 2.326
we got |zo| =8.252 & | z | =2.326
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -8.2524 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 < p2
test statistic: -8.252
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 =131, n1 =375, p1= x1/n1=0.349
sample two, x2 =244, n2 =375, p2= x2/n2=0.651
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( (0.349*0.651/375) +(0.651 * 0.349/375))
=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
=0.068
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.349-0.651) ±0.068]
= [ -0.37 , -0.233]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =131, n1 =375, p1= x1/n1=0.349
sample two, x2 =244, n2 =375, p2= x2/n2=0.651
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 = [ (0.349-0.651) ± 1.96 * 0.035]
= [ -0.37 , -0.233 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ -0.37 , -0.233] 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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