Question

Dorothy Kelly sells life insurance for the Prudence Insurance Company. She sells insurance by making visits to her clients homes. Dorothy believes that the number of sales should depend, to some degree, on the number of visits made. For the past several years, she kept careful records of the number of visits (x) she made each week and the number of people (y) who bought insurance that week. For a random sample of 15 such weeks, the x and y values follow.

x	11	21	18	15	28	5	20	14	22	7	15	29	8	25	16
y	1	10	9	4	8	2	5	6	8	3	5	10	6	10	7

In this setting we have Σx = 254, Σy = 94, Σx² = 5060, Σy² = 710, and Σxy = 1833.

(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 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 < 0.1250.075 < 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) In a week during which Dorothy makes 21 visits, how many people do you predict will buy insurance from her? (Round your answer to one decimal place.)
people

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

(g) Find a 95% confidence interval for the number of sales Dorothy would make in a week during which she made 21 visits. (Round your answers to one decimal place.)

lower limit	sales
upper limit	sales

(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 < 0.1250.075 < 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 an 80% confidence interval for β and interpret its meaning. (Round your answers to three decimal places.)

lower limit
upper limit

Interpretation

For each less visit made, sales increase by an amount that falls within the confidence interval.For each additional visit made, sales increase by an amount that falls within the confidence interval.    For each additional visit made, sales increase by an amount that falls outside the confidence interval.For each less visit made, sales increase by an amount that falls outside the confidence interval.

Answer 1

	n=	15.0000
	X̅=ΣX/n	16.9333
	Y̅=ΣY/n	6.2667
	sx=(√(Σx2-(Σx)2/n)/(n-1))=	7.3627
	sy=(√(Σy2-(Σy)2/n)/(n-1))=	2.9391
	Cov=sxy=(ΣXY-(ΣXΣY)/n)/(n-1)=	17.2333
	r=Cov/(Sx*Sy)=	0.796
	slope= β̂1 =r*Sy/Sx=	0.3179
	intercept= β̂0 ='y̅-β1x̅=	0.8835

a)

X̅=ΣX/n =	16.93
Y̅=ΣY/n =	6.27

b=0.3179

Least square line equation: ŷ =0.8835+0.3179*x

b)

c)

r=0.796

r² =0.634

63.4 %

d_)

t=r*(√(n-2)/(1-r2))=

4.748

P-value < 0.0005

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

e)

predicted val=0.884+21*0.318=

7.6

f)

Se =√(SSE/(n-2))=

1.845

g)

standard error of PI=s*√(1+1/n+(x0-x̅)2/Sxx)=		1.9245
for 95 % CI value of t =	2.160
margin of error E=t*std error   =		4.158
lower confidence bound=xo-E =		3.4
Upper confidence bound=xo+E=		11.7

h)

t=r*(√(n-2)/(1-r2))=

4.748

P-value < 0.0005

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

i)

std error of slope =se(β1) =s/√Sxx=		0.0670
for 90 % CI value of t =	1.771
margin of error E=t*std error   =		0.119
lower confidence bound=xo-E =		0.199
Upper confidence bound=xo+E=		0.436

.For each additional visit made, sales increase by an amount that falls within the confidence interval.