Use the given degree of confidence and sample data to construct interval for the

Question

Use the given degree of confidence and sample data to construct interval for the population proportion. Of 369 randomly selected medical students, 23 said that they planned to work in a rural community. Construct a 95% confidence interval for the percentage of all medical students who plan to work in a rural community. (4.16%, 8.30%) (3.77%, 9.47%) (3.30%, 9.17%) (2.99%, 9.47%) (3.77%, 8.70%) In a poll of registered voters nationwide, 43% of those polled blamed of companies the most for the recent increase in gasoline prices. The margin of error at the 96% confidence level for this point estimate is 2.4%. Construct a 95% confidence level for the population proportion who blame oil companies for the recent increase in gasoline prices. (0.382, 0.478) (0.368, 0.492) (0.406, 0.477) (0.383, 0.477) Cannot be determined from the information given.

Explanation / Answer

QUestion 8

p^ = 23/ 369 = 0.06233

95% confidenceinterval = p^ +- Z95% sqrt [p^ (1-p^)/N]

= 0.06233 +- 1.96 * sqrt [0.06233 * 0.9377/ 369]

= (3.77, 8.70)

Question 9

P^ = 43% = 0.43

Margin of errror = 2.4% = 0.024

95 % CI = p^ +- Z95% * Margin of error = 0.43 +- 1.96 * 0.024

= (0.383, 0.477) = (38.3%, 47.7%)

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