the student newspaper at a college e It would be larger because higher confidenc
ID: 3326859 • Letter: T
Question
the student newspaper at a college e It would be larger because higher confidence interval requifes d lai Bo e. We can't tell because it depends on the size of population e can't tell because the margin of error is random the county government is satisfied with 90% confidence, but she wants a smaller can we get a smaller margin of error, still with 90% confidence? 18. Another member of margin of error. How a. Take a larger sample, because larger samples result in smaller margins of eco b. Take a smaller sample, because smaller samples result in smaller margins of error c. Take another sample of the same size and you might be l d. Take a sample of adults from a neighborhood in the county instead of from population will be smaller which will be a smaller margin of error. lucky and get a much smaller margin of error the entire county . Then, the e Carry out a call-in poll to get a voluntary response sample. Voluntary response samples of error samples have no margin Use the following for questions 19-21: The student newspaper at a college asks a simple random sample of 200 undergrads eliminating supplemental fees for lab courses?" In all, 150 of the 200 are in favor of eliminating "Do you favor fees. 19. To estimate p, you will use the proportion a. bias b. confidence level c. mean d. parameter e. statistic 20. What is the 95% confidence interval for the population proportion p ? 21 , what happens if we take a 99.7% confidence interval instead? which one would you prefer? 22. A grocery chain runs a prize game by giving each customer a ticket that may win a prize wi is scratched. Printed on the ticket are the following probabilities for a customer who shops once Amount won $500 $150 Probability 0.025 0.05Explanation / Answer
Question 19
It would be a statistic as it is derived from a sample.
p = 150/200 = 0.75
Q.20
95% confidence interval = p^ +- Z95% sqrt [p^ * (1-p^)/N]
= 0.75 +- 1.96 * sqrt [0.75 * 0.25/200]
= 0.75 +- 1.96 * 0.03062
= (0.69, 0.81)
Question 21
99.7% confidence interval =
p^ +- Z99.7% sqrt [p^ * (1-p^)/N]
= 0.75 +- 3 * sqrt [0.75 * 0.25/200]
= 0.75 + 3 * 0.03062
= (0.658, 0.842)
The confidence interval would increase if we incease confidenceinterva.
we will prefere a 95% confidece interval.
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