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, 385 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." The study found that 66.7% of the 132 die-hard fans attended Cubs games at least once a month, but only 18.2% of the 253 less loyal fans attended this often. Analyze these data using a significance test for the difference in proportions. (Let

D = pdie-hard pless loyal. Use = 0.05. Round your value for z to two decimal places. Round your P-value to four decimal places.)

Analyze these data using a 95% confidence interval for the difference in proportions. (Round your answers to three decimal places.)

Explanation / Answer

PART 1

Given that,
sample one, n1 =132, p1= x1/n1=0.667
sample two, n2 =253, p2= x2/n2=0.182
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =
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 =(0.667-0.182)/sqrt((0.348*0.652(1/132+1/253))
zo =9.481
| zo | =9.481
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =9.481 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 9.4812 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 9.481
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0

PART II.

Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=88.044
No.Of Observed (n1)=132
P1= X1/n1=0.667
Proportion 2
No. of chances(X2)=46.046
No.Of Observed (n2)=253
P2= X2/n2=0.182
C.I = (0.667-0.182) ±Z a/2 * Sqrt( (0.667*0.333/132) + (0.182*0.818/253) )
=(0.667-0.182) ± 1.96* Sqrt(0.002)
=0.485-0.093,0.485+0.093
=[0.392,0.578]

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