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"The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study.
a. Provide a 95% confidence interval for the population proportion of college students who work to pay for tuition and living expenses. (to 2 decimals) ,
b. Provide a 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses. (to 2 decimals) ,
c. What happens to the margin of error as the confidence is increased from 95% to 99%?"
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Answer
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Step 1
To find the 95% confidence interval for the population proportion, we first need to calculate the standard error (SE), which is given by the formula: Plugging in the values, we get: Now, the 95% confidence interval is given by: (\hat{p} - z \times SE, \hat{p} + z \times SE) For a 95% confidence level, the z-score is approximately 1.96. Plugging in the values, we get: (0.47 - 1.96 \times 0.0302, 0.47 + 1.96 \times 0.0302) \approx (0.4108, 0.5292) So, the 95% confidence interval is (0.4108, 0.5292).
Step 2
To find the 99% confidence interval, we use the same formula as before, but with the z-score for the 99% confidence level, which is approximately 2.576. Plugging in the values, we get: (0.47 - 2.576 \times 0.0302, 0.47 + 2.576 \times 0.0302) \approx (0.3852, 0.5548) So, the 99% confidence interval is (0.3852, 0.5548).
Final Answer
a. The 95% confidence interval is (0.4108, 0.5292). b. The 99% confidence interval is (0.3852, 0.5548). c. As the confidence level increases from 95% to 99%, the margin of error increases, resulting in a wider confidence interval.
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