Researchers are interested in determining whether more men than women prefer Coca Cola to Pepsi. In a random sample of 300 men, 65% prefer Coca Cola, whereas in a random sample of 400 women, 48% prefer Coca Cola. What is the 99% confidence interval estimate for the difference between the percentages of men and women who prefer Coca Cola over Pepsi? 0.17 ± 0.036 0.17 ± 0.096 0.17 ± 0.067 0.565 ± 0.067 0.565 ± 0.096

0.17 ± 0.096

Step-by-step explanation:

Confidence interval for difference between the population proportion :

, where = sample proportion for population 1.

  = sample proportion for population 2.

= Sample size from population 1.

= Sample size from population 1.

As per given , we have

Population

Critical z-value for 99% confidence level is z*=2.576  [By z-table]

Now , the 99% confidence interval estimate for the difference between the percentages of men and women who prefer Coca Cola over Pepsi :

\

Hence, the correct answer is 0.17 ± 0.096.

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non-answer:

choosing

sorry if this isnt what you're looking for i dont really understand the question