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, 366 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 92 of the 133 die-hard fans attended Cubs games at least once a month, but only 41 of the 233 less loyal fans attended this often. Analyze these data using a significance test for the difference in proportions. (Let D = p_die-hard − p_{less loyal}. Use α = 0.05. Round your value for z to two decimal places. Round your P-value to four decimal places.)

z	=
P-value	=

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

( , )

Write a short summary of your findings.

1 Reject the null hypothesis, there is significant evidence that a higher proportion of die hard Cubs fans attend games at least once a month.

2 Reject the null hypothesis, there is not significant evidence that a higher proportion of die hard Cubs fans attend games at least once a month.   

3 Fail to reject the null hypothesis, there is not significant evidence that a higher proportion of die hard Cubs fans attend games at least once a month.

4 Fail to reject the null hypothesis, there is significant evidence that a higher proportion of die hard Cubs fans attend games at least once a month.

Answer 1

Hypothesis:
H0 : p1= p2
Ha: p1 not = p2

p1 = 92/133= 0.6917

p2 = 41/233 = 0.1760

pcap = (92 + 41)/(133+233) = 0.3634

test statistics:

z = (p1- p2)/sqrt(pcap *(1-pcap) *(1/n1+1/n2)
= (0.6917 - 0.1760)/sqrt(0.3634 *(1-0.3634)*(1/133+1/233))
= 9.87

p value = 0.0001

Here, , n1 = 133 , n2 = 233
p1cap = 0.6917 , p2cap = 0.176

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.6917 * (1-0.6917)/133 + 0.176*(1-0.176)/233)
SE = 0.0472

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.6917 - 0.176 - 1.96*0.0472, 0.6917 - 0.176 + 1.96*0.0472)
CI = (0.4232 , 0.6082)

1 Reject the null hypothesis, there is significant evidence that a higher proportion of die hard Cubs fans attend games at least once a month