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A Researcher Is Interested in Estimating the Proportion of Adults 0.003±1.645(0.003)(0.997)/10000.003 \pm 1.645 \sqrt { ( 0.003 ) ( 0.997 ) / 1000 }

Question 33

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A researcher is interested in estimating the proportion of adults in the U.S. who suffer from a rare form of cancer. In a random sample of 1000 adults in the U.S. she finds that 0.3% suffer from this form of cancer. She then obtains the following 90% confidence interval: 0.003±1.645(0.003)(0.997)/10000.003 \pm 1.645 \sqrt { ( 0.003 ) ( 0.997 ) / 1000 } or 0.000155 to 0.0058 She concludes that she can be 90% confident that the true proportion of adults in the U.S. suffering from this form of cancer is somewhere between 0.0155% and 0.58%. Is anything wrong with this reasoning? Explain your answer.

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In calculating a confidence interval, we...

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