Question

In: Statistics and Probability

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

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." Of the 134 die-hard fans, 91.0% reported that they had watched or listened to Cubs games when they were children. Among the 232 less loyal fans, 70.3% said that they watched or listened as children. (Let D = pdie-hardpless 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 α = 0.01. Round your z-value to two decimal places and your P-value to four decimal places.)

z=

P-value=

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

(    ,    )

Solutions

Expert Solution

SolutionA:

(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=

=0.91*132

=120.12

=120

less loyal fans=0.703*232=163.096=163

die-hard fans=120

less loyal fans=163

Solutionb:

z=(0.91-0.703)/sqrt((0.91*(1-0.91)/132)+(0.703*(1-0.703)/232))

z=5.31

P VALUE can be found in excel as

1-NORM.S.DIST(5.31;TRUE)

First in enter in one cell as

=NORM.S.DIST(5.31;TRUE)

=0.999999945

=1-0.999999945

= 5.5e-08

p=0.000000055

z=5.31

P=0.0000

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

z alpha/2 fro 95%=1.96

lower bound=(0.91-0.703)-1.96 sqrt(0.91*(1-0.91)/132+0.703*(1-0.703)/232)

=0.207-0.07643

=0.131

upper limit=(0.91-0.703)+1.96 sqrt(0.91*(1-0.91)/132+0.703*(1-0.703)/232)

=0.207+0.07643

=0.283

0.131,0.283


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