Question

Each box of Healthy Crunch breakfast cereal contains a coupon entitling you to a free package of garden seeds. At the Healthy Crunch home office, they use the weight of incoming mail to determine how many of their employees are to be assigned to collecting coupons and mailing out seed packages on a given day. (Healthy Crunch has a policy of answering all its mail on the day it is received.) Let x = weight of incoming mail and y = number of employees required to process the mail in one working day. A random sample of 8 days gave the following data.

x (lb)	12	23	14	6	12	18	23	25
y (Number of employees)	4	11	8	5	8	14	13	16

In this setting we have Σx = 133, Σy = 79, Σx² = 2527, Σy² = 911, and Σxy = 1490.

(a) Find x, y, b, and the equation of the least-squares line. (Round your 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) If Healthy Crunch receives 11 pounds of mail, how many employees should be assigned mail duty that day? (Round your answer to two decimal places.)

        employees

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

(g) Find a 95% for the number of employees required to process mail for 11 pounds of mail. (Round your answer to two decimal places.)

lower limit	employees
upper limit	employees

(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 pound of mail, the number of employees needed increases by an amount that falls within the confidence interval.

For each additional pound of mail, the number of employees needed increases by an amount that falls outside the confidence interval.

For each less pound of mail, the number of employees needed increases by an amount that falls within the confidence interval.

For each less pound of mail, the number of employees needed increases by an amount that falls outside the confidence interval.

Answer 1

a)

X̅=ΣX/n =	16.63
Y̅=ΣY/n =	9.88

b=0.5592

ŷ =	0.5789+0.5592x

c)

r=Cov/(Sx*Sy)=

0.869

coefficient of determination r2 =

0.755

percentage of variation in y is explained by the least-squares model=75.5%

d)

test statistic t =

r*(√(n-2)/(1-r2))=

4.296

0.0005 < P-value < 0.005

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

e)

predicted value =

6.73

f)

Se =√(SSE/(n-2))=

2.314

g)

lower =0.46

upper =13.00

h)

test statistic t =

r*(√(n-2)/(1-r2))=

4.296

0.0005 < P-value < 0.005

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

i)

lower =0.372

upper =0.747

For each additional pound of mail, the number of employees needed increases by an amount that falls within the confidence interval.