During one of the first \"beer wars\" in the early 1980s, a taste test between S

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; fifty-four 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 respective companies?

Explanation / Answer

n = 100    

p = 0.54    

% = 95    

Standard Error, SE = Ö{p(1 - p)/n} =    (0.54(1 - 0.54))/100 = 0.049839743

z- score = 1.959963985    

Width of the confidence interval = z * SE =     1.95996398454005 * 0.0498397431775085 = 0.0976841

Lower Limit of the confidence interval = P - width =     0.54 - 0.0976841016266425 = 0.4423159

Upper Limit of the confidence interval = P + width =     0.54 + 0.0976841016266425 = 0.6376841

The 95% confidence interval is [0.4423, 0.6377]

We are 95% confident that the true proportion of beer drinkers who preferred bud to Schlitz lies in the interval [0.4423, 0.6377]

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