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)	9	18	13	6	12	18	23	25
y (Number of employees)	7	8	10	5	8	14	13	16

In this setting we have Σx = 124, Σy = 81, Σx² = 2232, Σy² = 923, and Σxy = 1414.

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) If Healthy Crunch receives 17 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 17 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 > 0.2500.125 < 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 pound of mail, the number of employees needed increases by an amount that falls outside the confidence interval.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.

Answer 1

percentage of variation in y is explained by the least-squares model =r^2*100 =78.8 %

test statistic t =

r*(√(n-2)/(1-r2))=

4.719

0.0005 < P-value < 0.005

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

e)

predicted value =

10.89

f)

Se =√(SSE/(n-2))=

1.908

g)

std error of confidence interval =				s*√(1+1/n+(x0-x̅)2/Sxx)=			2.0299
for 95 % confidence and 6degree of freedom critical t=							2.447
lower limit =		5.92
uppr limit =		15.86

h)

test statistic t =

r*(√(n-2)/(1-r2))=

4.719

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=		0.1084
for 80 % confidence and -2degree of freedom critical t=						1.4400
80% confidence interval =b1 -/+ t*standard error=					(0.355,0.667)
lower limit =		0.355
uppr limit =		0.667

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