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5 January, 05:24

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?

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  1. 5 January, 07:30
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    99% confidence interval for the difference between the percentages of men and women who prefer Coca Cola over Pepsi is between a lower limit of 13.3% and an upper limit of 20.7%.

    Explanation:

    Confidence interval for a proportion is given as p + / - zsqrt[p (1-p) : n]

    difference between proportion (p) of men and women who prefer Coca Cola over Pepsi (p) = 0.65 - 0.48 = 0.17

    total number of men and women (n) = 300+400 = 700

    confidence level (C) = 99% = 0.99

    significance level = 1 - C = 1 - 0.99 = 0.01 = 1%

    critical value (z) at 1% significance level is 2.576.

    Error of margin = zsqrt[p (1-p) : n] = 2.576sqrt[0.17 (1-0.17) : 700] = 2.576 * 0.0142 = 0.037

    Lower limit of difference in proportion = 0.17 - 0.037 = 0.133 = 13.3%

    Upper limit of difference in proportion = 0.17 + 0.037 = 0.207 = 20.7%

    99% confidence interval is (13.3%, 20.7%)
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