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 20 15 12 28 5 20 14 22 7 15 29 8 25 16

y 3 12 10 5 8 2 5 6 8 3 5 10 6 10 7

In this setting we have Σx = 247, Σy = 100, Σx2 = 4839, Σy2 = 790, and Σxy = 1873.

(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 > 0.250

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) In a week during which Dorothy makes 20 visits, how many people do you predict will buy insurance from her? (Round your answer to one decimal place.)

_____ people

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

Se =

(g) Find a 95% confidence interval for the number of sales Dorothy would make in a week during which she made 20 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 > 0.250

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 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 within the confidence interval.

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 outside the confidence interval.

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

Answer 1

a)

X̅=ΣX/n =	16.47
Y̅=ΣY/n =	6.67

b=0.2933

ŷ =	1.8373+0.2933x

b)

scatter plot is attached above

c)

r =

0.734

coefficient of determination r2 =

0.538

d)

test statistic t =

r*(√(n-2)/(1-r2))=

3.892

0.0005 < P-value < 0.005

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

e)

predicted value =

7.70

f)

Se =√(SSE/(n-2))=

2.093

g)

for 95 % confidence and 13degree of freedom critical t=							2.160
95% confidence interval =yo -/+ t*standard error=					(3,12.4)

lower limit =3.1

upper limit =2.4

h)

test statistic t =

r*(√(n-2)/(1-r2))=

3.892

0.0005 < P-value < 0.005
Reject the null hypothesis. There is sufficient evidence that β > 0.

i)

lower limit=0.192

upper limit =0.395

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