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, 364 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 64.7% of the 136 die-hard fans attended Cubs games at least once a month, but only 17.5% of the 228 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

a.
Given that,
sample one, x1 =87.992, n1 =136, p1= x1/n1=0.647
sample two, x2 =39.9, n2 =228, p2= x2/n2=0.175
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 =(0.647-0.175)/sqrt((0.351*0.649(1/136+1/228))
zo =9.125
| zo | =9.125
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =9.125 & | 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 != 9.1254 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 9.125
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0

b.
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 )=87.992
No.Of Observed (n1)=136
P1= X1/n1=0.647
Proportion 2
No. of chances(X2)=39.9
No.Of Observed (n2)=228
P2= X2/n2=0.175
C.I = (0.647-0.175) ±Z a/2 * Sqrt( (0.647*0.353/136) + (0.175*0.825/228) )
=(0.647-0.175) ± 1.96* Sqrt(0.002)
=0.472-0.094,0.472+0.094
=[0.378,0.566]

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