Question

A sociologist is interested in the relation between x = number of job changes and y = annual salary (in thousands of dollars) for people living in the Nashville area. A random sample of 10 people employed in Nashville provided the following information.

x (number of job changes)	3	6	4	6	1	5	9	10	10	3
y (Salary in $1000)	35	36	38	32	32	38	43	37	40	33

In this setting we have Σx = 57, Σy = 364, Σx² = 413, Σy² = 13,364, and Σxy = 2143.

(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 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 5% 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 < 0.1250.075 < P-value < 0.1000.050 < P-value < 0.0750.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.0100.0005 < P-value < 0.005P-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 someone had x = 8 job changes, what does the least-squares line predict for y, the annual salary? (Round your answer to two decimal places.)
thousand dollars

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

(g) Find a 90% confidence interval for the annual salary of an individual with x = 8 job changes. (Round your answers to two decimal places.)

lower limit	thousand dollars
upper limit	thousand dollars

(h) Test the claim that the slope β of the population least-squares line is positive at the 5% 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 < 0.1250.075 < P-value < 0.1000.050 < P-value < 0.0750.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.0100.0005 < P-value < 0.005P-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 a 90% confidence interval for β and interpret its meaning. (Round your answers to three decimal places.)

lower limit
upper limit

Interpretation

For each less job change, the annual salary increases by an amount that falls within the confidence interval.For each less job change, the annual salary increases by an amount that falls outside the confidence interval.    For each additional job change, the annual salary increases by an amount that falls outside the confidence interval.For each additional job change, the annual salary increases by an amount that falls within the confidence interval.

Answer 1

a)

sample size ,   n =   10          
here, x̅ =Σx/n =   5.700   ,   ȳ = Σy/n =   36.400  

SSxx =    Σx² - (Σx)²/n =   88.10          
SSxy=   Σxy - (Σx*Σy)/n =   68.20          
SSyy =    Σy²-(Σy)²/n =   114.40          
estimated slope , ß1 = SSxy/SSxx =   68.200   /   88.100   =   0.77412
                  
intercept,   ß0 = y̅-ß1* x̄ =   31.98751          
                  
so, regression line is   Ŷ =   31.9875   +   0.7741   *x

b)

c)

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.679
      
R² =    (Sxy)²/(Sx.Sy) =    0.4615

d)

Ho:   ρ = 0  
Ha:   ρ > 0  
n=   10  
alpha,α =    0.05  
correlation , r=   0.6793  
t-test statistic = r*√(n-2)/√(1-r²) =        2.618
DF=n-2 =   8  
p-value =    0.0154  
Decison:   p value < α , So, Reject Ho  

0.010 < P-value < 0.025

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

e)

Predicted Y at X=   8   is                  
Ŷ =   31.988   +   0.774   *   8   =   38.180

f)

SSE=   (Sx*Sy - S²xy)/Sx =    61.6050
      
std error ,Se =    √(SSE/(n-2)) =    2.78

g)

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    1.110              
margin of error,E=t*Std error=t* S(ŷ) =   1.8595   *   1.1102   =   2.0644
                  
Confidence Lower Limit=Ŷ +E =    38.180   -   2.0644   =   36.12
Confidence Upper Limit=Ŷ +E =   38.180   +   2.0644   =   40.24

h)

t stat = estimated slope/std error =ß1 /Se(ß1) =    0.7741   /   0.2956   =   2.618

0.010 < P-value < 0.025

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

i)

confidence interval for slope                  
α=   0.1              
t critical value=   t α/2 =    1.860   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    2.77500   /√   88.10   =   0.296
                  
margin of error ,E= t*std error =    1.860   *   0.296   =   0.550
estimated slope , ß^ =    0.7741              
                  
                  
lower confidence limit = estimated slope - margin of error =   0.7741   -   0.550   =   0.224
upper confidence limit=estimated slope + margin of error =   0.7741   +   0.550   =   1.324
For each additional job change, the annual salary increases by an amount that falls within the confidence interval.