In: Statistics and Probability
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) | 4 | 3 | 6 | 6 | 1 | 5 | 9 | 10 | 10 | 3 |
| y (Salary in $1000) | 35 | 37 | 36 | 32 | 32 | 38 | 43 | 37 | 40 | 33 |
In this setting we have Σx = 57, Σy = 363, Σx2 = 413, Σy2 = 13,289, and Σxy = 2137.
(a)Find Se. (Round your answer to three decimal places.)
(b) Find a 90% confidence interval for the annual salary of an individual with x = 5 job changes. (Round your answers to two decimal places.)
1) lower limit
2) upper limit
(c) 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 =
(l) Find a 90% confidence interval for β and interpret its meaning. (Round your answers to three decimal places.)
1) lower limit
2) upper limit

| n= | 10.0000 | |
| X̅=ΣX/n | 5.7000 | |
| Y̅=ΣY/n | 36.3000 | |
| sx=(√(Σx2-(Σx)2/n)/(n-1))= | 3.1287 | |
| sy=(√(Σy2-(Σy)2/n)/(n-1))= | 3.5292 | |
| Cov=sxy=(ΣXY-(ΣXΣY)/n)/(n-1)= | 7.5444 | |
| r=Cov/(Sx*Sy)= | 0.6832 | |
| slope= β̂1 =r*Sy/Sx= | 0.7707 | |
| intercept= β̂0 ='y̅-β1x̅= | 31.9069 |
a)
| Se =√(SSE/(n-2))= | 2.733 |
b)
| predicted val=31.907+5*0.771= | 35.76 |
| standard error of PI=s*√(1+1/n+(x0-x̅)2/Sxx)= | 2.8740 | |
| for 90 % CI value of t = | 1.860 | |
| margin of error E=t*std error = | 5.344 | |
| lower confidence bound=xo-E = | 30.42 | |
| Upper confidence bound=xo+E= | 41.10 | |
c)
| t=r*(√(n-2)/(1-r2))= | 2.647 |
| critical t = | 1.860 |
d)
l)
| std error of slope =se(β1) =s/√Sxx= | 0.2912 | |
| for 90 % CI value of t = | 1.860 | |
| margin of error E=t*std error = | 0.542 | |
| lower confidence bound=xo-E = | 0.229 | |
| Upper confidence bound=xo+E= | 1.312 | |