Question

During one of the first "beer wars" in the early 1980s, a taste test between Schlitz and Budweiser was the focus of a nationally broadcast TV commercial. One hundred people agreed to drink from two unmarked mugs and indicate which of the two beers they liked better; fiftyfour said, "Bud." Construct and interpret the corresponding 95% confidence interval for p, the true proportion of beer drinkers who prefered Budweiser to Schlitz. How would Budweiser and Schlitz executives each have put these results in the best possible light for their perspective companies?

Answer 1

See Explaination

Explanation:

Looking at the question, The proportion of beer drinkers who preferred Budweiser to Schlitz as obtained from the sample is 54/100 = 0.54.

Let X be a random variable representing the number of drinkers who prefer Budweiser to Schlitz with total number of drinkers being n.

X \sim Bin(n, p)

From the sample, \hat{p} = 0.54

Therefore, the 95% confidence interval for p is given by,

(\hat{p}-z_{1-\alpha/2}\sqrt{\frac{1}{n}\hat{p}(1-\hat{p})}, \hat{p}+z_{1-\alpha/2}\sqrt{\frac{1}{n}\hat{p}(1-\hat{p})})

where z_{1-\alpha/2} is the upper \alpha/2 point of standard normal variate.

Here, \alpha = 0.05

Therefore, z_{1-\alpha/2} = z_{0.975} = 1.96

The required confidence interval for p is given by

(0.54 - 1.96\times \sqrt{\frac{0.54\times 0.46}{100}}, 0.54 + 1.96\times \sqrt{\frac{0.54\times 0.46}{100}}) = (0.442, 0.638)

Since the lower bound of the interval is less than 0.50 and upper bound is more than 0.50, therefore the two parties will use these bounds to glorify their brand. Budweiser will use the upper bound = 0.638 to say that the true proportion may be more than 0.5 indicating their beer is more preferred by the people. On the other hand, the lower bound = 0.442 will be used by the Schlitz to show that the true proportion may have values less than 0.5 indicating higher proportion of people favors their brand.