# 統計學的問題 關於母體樣本

5.There is concern about the speed of automobiles traveling over a particular of highway. For a random sample 28 automobiles,radar indicated the following speeds,in miles per hour:

59 63 68 57 56 71 59

69 53 58 60 66 51 59

54 64 58 57 66 61 65

70 63 65 57 56 61 59

Assuming a normal population distribution,find the margin of error of a 95% confidence interval for the mean speed of all automobiles traveling over this stretch of highway.

6.A clinic offers a weight-loss program. A review of its records found the following amounts of weight loss,inpounds,for a random sample of 24 of its clients at the conclusion of a 4-month program:

18 25 16 11 15 20 16 19

28 25 26 31 45 40 36 19

28 25 36 16 35 20 16 19

a.Find a 99% confidence interval for the population mean.

b.Without doing the calculations,explain whether a 90% confidence interval for the population mean would be wider than,narrower than,or the same as that found in part a.

7.In a random sample of 95 manufacturing firms,67indicated that their company attained ISO certification within the last two years. Find 99% confidence interval for the population proportion of companies that have been certified within the last 2 years.

8.Consider the following random sample from a normal population :

12 16 8 10 9

a.Find the 90% confidence interval for population variance.

b.Find the 95% confidence interval for the population variance.

9.A clinic offers a weight-loss program. A review of its records found the following amounts of weight loss,inpounds,for a random sample of 10 clients at the conclusion of the program:

18.2 25.9 6.3 11.8 15.4 20.3 16.8 18.5 12.3 17.2

Find a 90% confidence interval for the population variance of weight loss for clients of this weight-loss program.

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5.51 53 54 56 56 57 57 57 58 58 59 59 59 59 60 61 61 63 63 64 65 65 66 66 68 69 70 71 Sum=Σx=1705xbar=Σx/n=60.89285714S=Σ(x-xbar)/(n-1)=5.230532058σ=S/√n=0.988477646z(95%)=1.96Error margin=z*σ=1.937.....ans

6.a.Find a 99% confidence interval for the population mean.Ans:11 15 16 16 16 16 18 19 19 19 20 2025 25 25 26 28 28 31 35 36 36 40 45Sum=Σx=585xbar=Σx/n=24.375S=Σ(x-xbar)/(n-1)=8.943360423σ=s/√n=1.825555802z(99%)=2.575μ=xbar+-z*σ=24.375+-2.575*1.83=19.66~29.09.....ans

6.b.z(90%)=1.645μ=xbar+-z*σ=24.375+-1.645*1.83=21.36~27.39.....ans=Narrower than (a)

7.p=67/95=0.7053q=1-p=0.2947σ=√(pq/n)=√(0.7053*0.2947/95)=0.04678z(99%)=2.575p^=p+-z*σ=0.7053+-2.575*0.04678=0.585~0.826.....ans

8.a.8 9 10 12 16sum=Σx=55xbar=Σx/n=11S=Σ(x-xbar)/(n-1)=3.16227766σ=S/√n=1.414213562z(90%)=1.645variance interval=z*σ=1.645*1.4142=2.326.....ans

8.b.z(95%)=1.96variance interval=z*σ=1.96*1.4142=2.77.....ans

9.6.3 11.8 12.3 15.4 16.8 17.2 18.2 18.5 20.3 25.9Sum=Σx=162.7xbar=Σx/n=16.27S=Σ(x-xbar)/(n-1)=5.319993734σ=S/√n=1.682329734z(90%)=1.645variance interval=z*σ=1.645*1.682=2.767.....ans