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)	5	3	4	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 = 56, Σy = 363, Σx² = 402, Σy² = 13,289, and Σxy = 2100.

(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.

11

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 someone had x = 6 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 = 6 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 > 0.250

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 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

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

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

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

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

PLEASE NOTE THIS IS ALL ONE QUESTION. ALL ONE QUESTION. Thanks.

Answer 1

c)

r=0.675

r² =0.456

45.6%

d)

test statistic t =

r*(√(n-2)/(1-r2))=

2.588

0.010 < P-value < 0.025

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

e)

predicted value =

36.60

f)

Se =√(SSE/(n-2))=

2.76

g_

std error of confidence interval =				s*√(1+1/n+(x0-x̅)2/Sxx)=			2.8989
for 90 % confidence and 8degree of freedom critical t=							1.860
lower limit =		31.21
uppr limit =		41.99

h)

test statistic t =

r*(√(n-2)/(1-r2))=

2.588

0.010 < P-value < 0.025  

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

i)

std error of slope sb1 =				s/√SSx=		0.2937
for 90 % confidence and -2degree of freedom critical t=						1.8600
90% confidence interval =b1 -/+ t*standard error=					(0.214,1.307)
lower limit =		0.214
uppr limit =		1.307

D)For each additional job change, the annual salary increases by an amount that falls within the confidence interva