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Q1: A random sample of 390 married couples found that 290 had two or more personality...

Q1: A random sample of 390 married couples found that 290 had two or more personality preferences in common. In another random sample of 570 married couples, it was found that only 36 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common.

(a) Find a 95% confidence interval for p1p2. (Use 3 decimal places.)

lower limit = ?

upper limit = ?

Q2: Inorganic phosphorous is a naturally occurring element in all plants and animals, with concentrations increasing progressively up the food chain (fruit < vegetables < cereals < nuts < corpse). Geochemical surveys take soil samples to determine phosphorous content (in ppm, parts per million). A high phosphorous content may or may not indicate an ancient burial site, food storage site, or even a garbage dump. Independent random samples from two regions gave the following phosphorous measurements (in ppm). Assume the distribution of phosphorous is mound-shaped and symmetric for these two regions

Region I: x1; n1 = 15
855 1550 1230 875 1080 2330 1850 1860
2340 1080 910 1130 1450 1260 1010
Region II: x2; n2 = 14
540 810 790 1230 1770 960 1650 860
890 640 1180 1160 1050 1020

(a) Use a calculator with mean and standard deviation keys to verify that x1, s1, x2, and s2. (Round your answers to one decimal place.)

x1 = ___ ppm
s1 = ___ ppm
x2 = ___ ppm
s2 = ___ ppm


(b) Let ?1 be the population mean for x1 and let ?2 be the population mean for x2. Find a 98% confidence interval for ?1 ? ?2. (Round your answers to one decimal place.)

lower limit ___ ppm
upper limit ___ ppm

Q3: For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

In a survey of 1000 large corporations, 260 said that, given a choice between a job candidate who smokes and an equally qualified nonsmoker, the nonsmoker would get the job.

(a) Let p represent the proportion of all corporations preferring a nonsmoking candidate. Find a point estimate for p. (Round your answer to four decimal places.)

(b) Find a 0.95 confidence interval for p. (Round your answers to three decimal places.)

lower limit = ?

upper limit = ?


Q4: Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 12 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with ? = 0.20 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)

lower limit = ?
upper limit = ?
margin of error = ?

(b) What conditions are necessary for your calculations? (Select all that apply.)? is unknownnormal distribution of weightsuniform distribution of weights? is knownn is large

Q5: Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of 42 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that ? = 8.00 ml/kg for the distribution of blood plasma.

(a) Find a 99% confidence interval for the population mean blood plasma volume in male firefighters. What is the margin of error? (Round your answers to two decimal places.)

lower limit = ?
upper limit = ?
margin of error = ?

Homework Answers

Answer #2

Solution1;

let p1^ be sample proportion who had two or more personality preferences in common

and p2^ be  sample proportion who had no preferences in common.

p1^=x1/n1=290/390=0.7435897

p2^=x2/n2=36/570=0.06315789

Z crit=95%=1.96

95% confidence interval for diff in proportions

=p1^-p2^+-Z sqrt(p1^(1-p1^)/n1+p2^(1-p2^)/n2

=0.7435897-0.06315789+-1.96sqrt(0.7435897*(1-0.7435897)/390+0.06315789*(1-0.06315789)/570)

=0.6804318+-0.04772

0.6804318-0.04772,0.6804318+0.04772

=0.633,0.728

lower limit=0.633

upper limit=0.728

0.633<p1-p2<0.728

we are 95% confident that the true population proportion difference lies in between 0.633 and 0.728

answered by: anonymous
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