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, 365 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 65.7% of the 137 die-hard fans attended Cubs games at least once a month, but only 19.3% 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.)

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Explanation / Answer

So, we're performing a two-proportion z test here because it's asking for the difference.Since we're assuming that the standard error is the same, we're going to have to pool the values.

.657 * 137 = 90, .193 * 228 = 44

I found the actual number of people in each category of fans that attended the games. Now, I'll pool them.

ppool = (90 + 44) / (137 + 228) = 0.3671

So now we need to find our standard error of the two proportions.

SE(p1 - p2) = sqrt[(0.3671 * 0.6329) / 137 + (0.3671 * 0.6329) / 228] = 0.0521

Now we do our z test. The 0.6329 came from 1 - ppool.

z = (0.657 - 0.193) / 0.0521 = 8.90 <===Extremely high!

P-value from 8.90
We can definitely reject the null because that number way below an alpha of 0.05.

The z* for 95% confidence is 1.96. We simply take the difference of the proportions and add/subtract the z* times the standard error.

0.464 +/- 1.96 * 0.0521 => (0.361, 0.566)

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