Question

Modern medical practice tells us not to encourage babies to become too fat. Is there a positive correlation between the weight x of a 1-year old baby and the weight y of the mature adult (30 years old)? A random sample of medical files produced the following information for 14 females.

x (lb)	19	24	24	26	20	15	25	21	17	24	26	22	18	19
y (lb)	122	120	125	130	130	120	145	130	130	130	130	140	110	115

In this setting we have Σx = 300, Σy = 1777, Σx² = 6590, Σy² = 226,659, and Σxy = 38,288.

(a) Find x, y, b, and the equation of the least-squares line. (Round your answers for x and y to two decimal places. Round your answers for least-squares estimates to four decimal places.)

x	=
y	=
b	=
ŷ	=	+  x

(b) Draw a scatter diagram displaying the data. Graph the least-squares line on your scatter diagram. Be sure to plot the point (x, y).

(c) Find the sample correlation coefficient r and the coefficient of determination. (Round your answers to three decimal places.)

r =
r2 =

What percentage of variation in y is explained by the least-squares model? (Round your answer to one decimal place.)
%

(d) Test the claim that the population correlation coefficient ρ is positive at the 1% level of significance. (Round your test statistic to three decimal places.)

t =

Find or estimate the P-value of the test statistic.

P-value > 0.250

0.125 < P-value < 0.250  

  0.100 < P-value < 0.125

0.075 < P-value < 0.100

0.050 < P-value < 0.075

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

0.0005 < P-value < 0.005

P-value < 0.0005

Conclusion

Reject the null hypothesis. There is sufficient evidence that ρ > 0.Reject the null hypothesis. There is insufficient evidence that ρ > 0.    Fail to reject the null hypothesis. There is sufficient evidence that ρ > 0.Fail to reject the null hypothesis. There is insufficient evidence that ρ > 0.

(e) If a female baby weighs 25 pounds at 1 year, what do you predict she will weigh at 30 years of age? (Round your answer to two decimal places.)
lb

(f) Find S_e. (Round your answer to two decimal places.)
S_e =

(g) Find a 95% confidence interval for weight at age 30 of a female who weighed 25 pounds at 1 year of age. (Round your answers to two decimal places.)

lower limit	lb
upper limit	lb

(h) Test the claim that the slope β of the population least-squares line is positive at the 1% level of significance. (Round your test statistic to three decimal places.)

t =

Find or estimate the P-value of the test statistic.

P-value > 0.2500.

125 < P-value < 0.250   

0.100 < P-value < 0.125

.075 < P-value < 0.100

0.050 < P-value < 0.075

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

0.0005 < P-value < 0.005

P-value < 0.0005

Conclusion:

Reject the null hypothesis. There is sufficient evidence that β > 0.

Reject the null hypothesis. There is insufficient evidence that β > 0.    

Fail to reject the null hypothesis. There is sufficient evidence that β > 0.

Fail to reject the null hypothesis. There is insufficient evidence that β > 0.

(i) Find an 80% confidence interval for β and interpret its meaning. (Round your answers to three decimal places.)

lower limit ?
upper limit ?

Interpretation:

For each pound more a female infant weighs at 1 year, the adult weight increases by an amount that falls outside the confidence interval.

For each pound less a female infant weighs at 1 year, the adult weight increases by an amount that falls outside the confidence interval.

For each pound more a female infant weighs at 1 year, the adult weight increases by an amount that falls within the confidence interval.

For each pound less a female infant weighs at 1 year, the adult weight increases by an amount that falls within the confidence interval.

Answer 1

a)

total sum	300.000	1777.000	38288.00	6590.000	226659
mean	21.4286	126.9286

sample size ,   n =   14          
here, x̅ =Σx/n =   21.4286   ,   ȳ = Σy/n =   126.9285714  
                  
SSxx =    Σx² - (Σx)²/n =   161.429          
SSxy=   Σxy - (Σx*Σy)/n =   209.429          
SSyy =    Σy²-(Σy)²/n =   1106.929          
estimated slope , ß1 = SSxy/SSxx =   209.429   /   161.429   =   1.2973
                  
intercept,   ß0 = y̅-ß1* x̄ =   99.1283          
                  
so, regression line is   Ŷ =   99.1283   +   1.2973   *x

.................

b)

c)

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.4954
      
R² =    (Sxy)²/(Sx.Sy) =    0.2455

24.55% of variation in y is explained by the least-squares model

............

d)

correlation hypothesis test      
Ho:   ρ = 0  
Ha:   ρ ╪ 0  
n=   14  
alpha,α =    0.01  
correlation , r=   0.4954  
t-test statistic = r*√(n-2)/√(1-r²) =        1.976
DF=n-2 =   12  
p-value =    0.0716  
Decison:   P value > α, So, Do not reject Ho  
.Fail to reject the null hypothesis. There is insufficient evidence that ρ > 0.

.............

e)

Predicted Y at X=   25   is                  
Ŷ =   99.12832   +   1.297345   *   25   =   131.56

f)

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    835.227
      
std error ,Se =    √(SSE/(n-2)) =    8.343

.......

g)

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    3.236              
margin of error,E=t*Std error=t* S(ŷ) =   2.1788   *   3.2359   =   7.0505
                  
Confidence Lower Limit=Ŷ +E =    131.562   -   7.050   =   124.51
Confidence Upper Limit=Ŷ +E =   131.562   +   7.050   =   138.61

...........

h)

slope hypothesis test               tail=   2
Ho:   ß1=   0          
H1:   ß1╪   0          
n=   14              
alpha =   0.01              
estimated std error of slope =Se(ß1) = Se/√Sxx =    8.343   /√   161.43   =   0.6566
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    1.2973   /   0.6566   =   1.9758
                     
Degree of freedom ,df = n-2=   12              
p-value =    0.0716              
decison :    p-value>α , do not reject Ho              
Conclusion:   do not Reject Ho and conclude that slope is not significanty different from zero         

...........

i)

confidence interval for slope                  
α=   0.2              
t critical value=   t α/2 =    1.356   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    8.34280   /√   161.43   =   0.657
                  
margin of error ,E= t*std error =    1.356   *   0.657   =   0.891
estimated slope , ß^ =    1.2973              
                  
                  
lower confidence limit = estimated slope - margin of error =   1.2973   -   0.891   =   0.4068
upper confidence limit=estimated slope + margin of error =   1.2973   +   0.891   =   2.1879



For each pound more a female infant weighs at 1 year, the adult weight increases by an amount that falls within the confidence interval.

thanks

revert back for dount

please upvote