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	13	18	18	12	28	5	20	14	22	7	15	29	8	25	16
y	4	13	8	6	8	2	5	6	8	3	5	10	6	10	7

In this setting we have Σx = 250, Σy = 101, Σx² = 4910, Σy² = 797, and Σxy = 1892.

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

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 23 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 23 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 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.    For each additional visit made, sales increase by an amount that falls within the confidence interval.For each less visit made, sales increase by an amount that falls within the confidence interval.

Answer 1

percentage of variation in y is explained by the least-squares model =r²*100 =50.1 %

test statistic t =

r*(√(n-2)/(1-r2))=

3.612

0.0005 < P-value < 0.005

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

e)

predicted value =

8.51

f)

Se =√(SSE/(n-2))=

2.119

g)

std error of confidence interval =				s*√(1+1/n+(x0-x̅)2/Sxx)=			2.2428
for 95 % confidence and 13degree of freedom critical t=							2.160
lower limit =		3.7
uppr limit =		13.4

h_)

test statistic t =

r*(√(n-2)/(1-r2))=

3.612

0.0005 < P-value < 0.005

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

i)

std error of slope sb1 =				s/√SSx=		0.0777
for 80 % confidence and -2degree of freedom critical t=						1.3500
80% confidence interval =b1 -/+ t*standard error=					(0.176,0.386)
lower limit =		0.176
uppr limit =		0.386

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