BELLA 發問時間： 科學數學 · 8 年前

# 急~~生物統計學原理解答 ""20點""

1.Is a person’s body height predictive of his/her weight?Data on weight(y) in kilograms and height(x) in centimeters on 5 college students is contained.

Height weight

162 57

164 62

167 64

170 64

172 68

1.Report the least-squares fitted model.

2.Explain the coefficient of explanatory variable and intercept.

3.What is the estimated mean weight for the population of college students whose height is 172 centimeters?

4.Calculate the standard deviation of regression.

5.Test the null hypothesis that the true population slop β=0.

6.Construct a 95% confidence intervals for the coefficient of height.

7.Construct a 95% confidence intervals for the weight of a randomly selected a college studen whose height is 172 cm.

2.已知人體的血壓值與日常攝取的鹽量有關，下表是5個人的舒張壓及日常攝取鹽量

1 / 2 / 3 / 4 / 5

1.請以最小平方法求出Ŷ=a+bx的線性迴歸模式，其檢定斜率是否為0?

2.請計算迴歸係數b的95%CI。

3.由(2)所建立之迴歸模式，請解釋迴歸係數b的意義。

4.請計算由(2)所建立之迴歸模式的判定係數，並解釋所獲得數據的意義。

5.推測當日常攝取量為10g時，平均的舒張壓及其95%CI是多少?

6.已知小名的日常攝鹽量為12g，請預測小名的舒張壓及其95%CI是多少?

### 1 個解答

• 麻辣
Lv 7
8 年前
最佳解答

1.Is a person's body height predictive of his/her weight? Data on weight(y) in kilograms and height(x) in centimeters on 5 college students is contained.(一個人的身高&體重是否可以預測出來.5名大專學生的身重如下所示:)x///////x^2/////y///////x*y

162/////26244///57//////9234

164/////26896///62//////10168

167/////27889///64//////10688

170/////28900///64//////10880

172/////29584///68//////11696

835/////139513//315/////526661.Report the least-squares fitted model.(計算最少平方法模式)y=m*x+b, n=5Sx=ΣXj=sum(a1:a5)=835...........sum=Excel函數Sy=ΣYj=sum(c1:c5)=315Sx2=ΣXj^2=sum(b1:b5)=139513Sxy=Σ(Xj*Yj)=sum(d1:d5)=52666依據統計學線形回歸公式:m=(n*Sxy-Sx*Sy)/(n*Sx2-Sx^2)=0.897b=(Sy-m*Sx)/n=-86.8∴y=0.897x-86.8

2.Explain the coefficient of explanatory variable and intercept.(說明變數係數與截距)Slope=m=0.897截距=b=-86.8

3.What is the estimated mean weight for the population of college students whose height is 172 centimeters?(求172cm的體重)y(172)=0.897*172-86.8=67.5(kg)

4.Calculate the standard deviation of regression.(求回歸標準差)Xbar=[y(162)+y(164)+y(167)+y(170)+y(172)]/5=(58.5+60.3+63+65.7+67.5)/5=63S=√[(58.5^2+60.3^2+63^2+65.7^2+67.5^2)/5-63^2]=3.31.......ans

5.Test the null hypothesis that the true population slope β=0.(測試虛無假設slope=0的人數)ans=0

6.Construct a 95% confidence intervals for the coefficient of height.(建立身高係數.CI=95%)S=stdevp(a1:a5)=3.69Xbar=average(a1:a5)=167Xbar=167+-1.96*3.69=160~174

7.Construct a 95% confidence intervals for the weight of a randomly

selected a college studen whose height is 172 cm.(建立172cm體重的機率.CI=95%)Z=(67.5-63)/(3.31/√(n-1))=4.5*2/3.31=2.72P(0<Z<2.72)=0.4967P(Z<=2.72)=0.4967+0.5=0.9967

2.已知人體的血壓值與日常攝取的鹽量有關，下表是5個人的舒張壓及日常攝取

請答出解答和詳細解答過程喔~~謝謝><樣本:........1 / 2 / 3 / 4 / 5攝取量(g)(X) 15 /13 /11 /12 /9 舒張壓(mmhg)(Y)101/98/85/91/781.請以最小平方法求出Y=a+bx的線性迴歸模式，其檢定斜率是否為0?a.在Excel裡面操作b.A欄鍵入資料:15.13.11.12.9c.B欄鍵入資料:101.98.85.91.78d.鍵入=slope(b1:b5,a1:a5)=4.1=\=0e.鍵入=intercept(b1:b5,a1:a5)=41.4y=4.1x+41.4.......ans

2.請計算迴歸係數b的95%CI。b=41.4

2013-04-02 09:35:41 補充：

字數超過2000

不能再張貼

請繼續參考意見欄

2013-04-02 09:36:18 補充：

3.由(2)所建立之迴歸模式，請解釋迴歸係數b的意義。

b=y截距=41.4........ans

4.請計算由(2)所建立之迴歸模式的判定係數，並解釋所獲得數據的意義。

相關係數=r=correl(a1:a5,c1:c5)=0.925

判定係數=r^2=0.856 => 高度相關

2013-04-02 09:36:41 補充：

5.推測當日常攝取量為10g時，平均的舒張壓及其95%CI是多少?

y(10)=4.1*10+41.4=82.4......ans

6.已知小名的日常攝鹽量為12g，請預測小名的舒張壓及其95%CI是多少?

y(12)=4.1*12+41.4=90.6.......ans