A health care professional wishes to estimate the birth weights of infants. How

Question

A health care professional wishes to estimate the birth weights of infants. How large a sample must be obtained if he desires to be 99% confident that he true mean is within 2 ounces of the sample mean and he knows that the population variance from past studies is 64 ounces? Answer __________ Interpreting a confidence interval: A magazine reports survey results of 500 randomly selected adults by stating that "38% of those surveyed favor Donald Trump for president with a margin of error of plusminus 7.5%." What confidence interval is suggested by the statement? Answer (Interval) __________ It is believed that the majority of Americans own pets. A 90% confidence interval for the percentage of Americans who owned pets is between 48 and 58 percent. What is the sample size? Answer __________

Explanation / Answer

Q.8 Confidence interval = 99%

Population variance 2 = 64 ounces

standard deviation = 8 ounces

so, Margin of error = 2 ounces = Z99% (/n)

2 = 2.55 * (8/ n)

n = 2.55 * 8/2 = 10.2

n = 104.04 or 104

Q.9 N = 500 ; p^ = 0.38

se0 = sqrt [ p^ * (1-p^) /N]

Margin of error = Za se0

0.075 = Za * sqrt [ 0.38 * 0.62/ 500]

Za = 0.075/ 0.0217 = 3.456

COnfidence interval = 0.001 Here

Q.10 HEre 90% confidence interval for proportion= (0.48, 0.58)

so p^ = (0.58 + 0.48)/2 = 0.53

so 0.58 - 0.48 = Z90% * sqrt [p* (1-p)/N]

0.10 = 1.645 * sqrt [0.53 * 0.47/N]

sqrt [0.53 * 0.47/N] = 0.06079

N = 67.40 or 68

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