Many species of cuckoos are brood parasites. They lay their eggs in the nests of
ID: 3431312 • Letter: M
Question
Many species of cuckoos are brood parasites. They lay their eggs in the nests of other species. The following data are for the lengths (mm) of cuckoo eggs in a random sample of nests from two different host species.
wren: 19.8, 22.1, 21.5, 20.9, 22, 21, 22.3, 21, 20.3, 20.9, 22, 20, 20.8, 21.2, 21
house sparrow: 22, 23.9, 20.9, 23.8, 25, 24, 21.7, 23.8, 22.8, 23.1, 23.1, 23.5, 23, 23
Question 7 (1 point)
Calculate a 95% confidence interval for the mean length of cuckoo eggs for each of the host species.
Question 7 options:
wren (20.3, 22.0), house sparrow (22.0, 23.9)
wren (20.7, 21.5), house sparrow (22.6, 23.7)
wren (19.6, 22.6), house sparrow (21.0, 25.2)
wren (20.9, 21.3), house sparrow (22.8, 23.4)
wren (20.3, 22.0), house sparrow (22.0, 23.9)
wren (20.7, 21.5), house sparrow (22.6, 23.7)
wren (19.6, 22.6), house sparrow (21.0, 25.2)
wren (20.9, 21.3), house sparrow (22.8, 23.4)
Many species of cuckoos are brood parasites. They lay their eggs in the nests of other species. The following data are for the lengths (mm) of cuckoo eggs in a random sample of nests from two different host species. wren: 19.8, 22.1, 21.5, 20.9, 22, 21, 22.3, 21, 20.3, 20.9, 22, 20, 20.8, 21.2, 21 house sparrow: 22, 23.9, 20.9, 23.8, 25, 24, 21.7, 23.8, 22.8, 23.1, 23.1, 23.5, 23, 23 Question 7 (1 point) Calculate a 95% confidence interval for the mean length of cuckoo eggs for each of the host species. Question 7 options: wren (20.3, 22.0), house sparrow (22.0, 23.9) wren (20.7, 21.5), house sparrow (22.6, 23.7) wren (19.6, 22.6), house sparrow (21.0, 25.2) wren (20.9, 21.3), house sparrow (22.8, 23.4)Explanation / Answer
wren: n1=15, xbar1=21.12, s1=0.7542262
The degree of freedom =n-1=15-1=14
Given a=0.05, t(0.025, df=14) =2.14 (from standard normal table)
So the lower bound is
xbar - t*s/vn =21.12 - 2.14*0.7542262/sqrt(15) =20.70326
So the upper bound is
xbar + t*s/vn =21.12 + 2.14*0.7542262/sqrt(15) =21.53674
house sparrow: n2=14, xbar2=23.11429, s2=1.049437
The degree of freedom =n-1=14-1=13
t(0.025, df=13) =2.16 (from student t table)
So the lower bound is
xbar - t*s/vn =23.11429 - 2.16*1.049437/sqrt(14) =22.50847
So the upper bound is
xbar + t*s/vn =23.11429 + 2.16*1.049437/sqrt(14) =23.72011
Answer: wren (20.7, 21.5), house sparrow (22.6, 23.7)
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