9. Find the margin of error for 99% confidence level for the population mean pri
ID: 3359636 • Letter: 9
Question
9. Find the margin of error for 99% confidence level for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. Round your answer to the nearest cent. (Reference problem 9) a) s0.65 per 100 pounds b) 13.47 per 100 pounds c) s0.25 per 100 pounds d) s0.07 per 100 pounds e) s0.19 per 100 pounds 10. In a random sample of 61 professional actors, it was found that 53 were extroverts. Find a 95% confidence interval for p. a) 0.782 to 0.955 b) 0.797 to 0.955 c) 0.797 to 0.941 d) 0.782 to 1.433 e) 0.782 to 0.941 Problem 1 A study shows that the proportion of Hispano or Latino in Florida during 2012 was 23.2%. Assuming that the sample selected was of 5,500 residents of Florida. Construct a 95% confidence interval. Source:http://quickfacts.census.gov/qfd/states/12000.html 11. The best point estimate of the population proportion is a. p=.0232 b, = .0.232 c, p=.0232 d. =0.232 12. The standard error,qslaun a. 0.0000323 b. 0.005692 c. 0.00000412 d. 0.00203Explanation / Answer
10.
p( extroverts) = x/n = 53/61 = .869
The Z value for 95% is ~2
Therefore the 95% CI is : p +/- Z*sqrt(p*p/n) = 0.869+/-2*SQRT(0.869*0.131/61)
= .782 to .955
Answer is A)
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