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    An insurance company is thinking about offering discounts on its life insurance policies to nonsmokers. As part of its analysis it randomly selects 200 men who are 60 years old and asks them whether or not they smoke at least one pack of cigarettes per day and whether they have ever suffered from heart disease. The results are stored in file smokers, where 2= suffer form heart disease and 1= do not suffer from heart disease.

    a. Can the company conclude at the 10% significance level that smokers have higher incidence of heart disease than nonsmokers?
    b. Estimate with 90% confidence the difference in the fractions of men suffering from heart disease between smokers and nonsmokers.

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    a. Can the company conclude at the 10% significance level that smokers have higher incidence of heart disease than nonsmokers?
    <br>
    <br>There are 38 smokers and 162 non-smokers.
    <br>Within the smokers, 10 persons suffer form heart disease, and the proportion is p1 = 10/38 = 26.32%
    <br>Within the non-smokers, 12 persons suffer form heart disease, and the proportion is p1 = 12/162 = 7.41%
    <br>the difference between the sample proportions is
    <br>DP = p1 - p2 = 0.2632 - 0.0741 = 0.1891.
    <br>
    <br>Then we are to ...

    Solution Summary

    An insurance company is thinking about offering discounts on its life insurance policies to nonsmokers. As part of its analysis it randomly selects 200 men who are 60 years old and asks them whether or not they smoke at least one pack of cigarettes per day and whether they have ever suffered from heart disease. The results are stored in file smokers, where 2= suffer form heart disease and 1= do not suffer from heart disease.

    a. Can the company conclude at the 10% significance level that smokers have higher incidence of heart disease than nonsmokers?
    b. Estimate with 90% confidence the difference in the fractions of men suffering from heart disease between smokers and nonsmokers.

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