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,
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.) Letx= weightof incoming mail andy= number of employees required to process the mail in one working day. A random sample of 8 days gave the following data.
x(lb) | 10 | 20 | 15 | 6 | 12 | 18 | 23 | 25 |
y(Number of employees) | 4 | 9 | 11 | 5 | 8 | 14 | 13 | 16 |
In this setting we havex=129,y=80,x2=2383,y2=928, andxy=1462.(a) Findx,y,b, and the equation of the least-squares line. (Round yourxandyto two decimal places. Round your least-squares estimates to four decimal places.)
x | = | |
y | = | |
b | = | |
= | + 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 coefficientrand the coefficient of determination. (Round your answers to three decimal places.)
r= | |
r2= |
What percentage of variation inyis explained by the least-squares model? (Round your answer to one decimal place.) % (d) Test the claim that the population correlation coefficientis positive at the 1% level of significance. (Round your test statistic to three decimal places.)
t=
Find or estimate theP-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> 0.
Reject the null hypothesis, there is insufficient evidence that> 0.
Fail to reject the null hypothesis, there is sufficient evidence that> 0
.Fail to reject the null hypothesis, there is insufficient evidence that> 0.
(e) If Healthy Crunch receives20pounds of mail, how many employees should be assigned mail duty that day? (Round your answer to two decimal places.) employees (f) FindSe. (Round your answer to three decimal places.) Se= (g) Find a 95% for the number of employees required to process mail for20pounds of mail. (Round your answer to two decimal places.)
lower limit | employees |
upper limit | employees |
(h) Test the claim that the slopeof 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 theP-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> 0.
Reject the null hypothesis, there is insufficient evidence that> 0.
Fail to reject the null hypothesis, there is sufficient evidence that> 0.
Fail to reject the null hypothesis, there is insufficient evidence that> 0. (i) Find an 80% confidence interval forand 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 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 additional 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.
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