Question

Bardi Trucking Co., located in Cleveland, Ohio, makes deliveries in the Great Lakes region, the Southeast, and the Northeast. Jim Bardi, the president, is studying the relationship between the distance a shipment must travel and the length of time, in days, it takes the shipment to arrive at its destination. To investigate, Mr. Bardi selected a random sample of 20 shipments made last month. Shipping distance is the independent variable and shipping time is the dependent variable. The results are as follows:

Shipment		Distance (miles)	Shipping Time (days)		Shipment	Distance (miles)	Shipping Time (days)
1		808	3		11	618	4
2		622	12		12	634	13
3		618	8		13	667	7
4		786	8		14	638	6
5		776	9		15	844	15
6		838	3		16	783	14
7		620	6		17	756	9
8		751	11		18	840	10
9		780	11		19	754	13
10		649	7		20	742	15

Click here for the Excel Data File

1. Draw a scatter diagram. Based on these data, does it appear that there is a relationship between how many miles a shipment has to go and the time it takes to arrive at its destination?

1. On the graph below, use the point tool to plot the point corresponding to the first Distance and its Shipping Time (Distance 1).

2. Repeat the process for the remainder of the sample (Distance 2, Distance 3, … ).

3. To enter exact coordinates, double-click on the point and enter the exact coordinates of x and y.

1. b-1. Fill in the blanks. (Round your answers to 3 decimal places. Negative values should be indicated by minus sign.)

x¯x¯
y¯y¯
Sx
Sy
r

1. b-2. State the decision rule for 0.05 significance level: H₀: ρ ≤ 0; H₁: ρ > 0. (Round your answer to 3 decimal places.)

1. b-3. Compute the value of the test statistic. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

1. b-4. Can we conclude that there is a positive correlation between distance and time? Use the 0.05 significance level.

1. c-1. Determine the coefficient of determination. (Round your answer to 3 decimal places.)

1. c-2. Fill in the blank below. (Round your answer to 1 decimal place.)

4. Determine the standard error of estimate. (Round your answer to 3 decimal places.)

5. Would you recommend using the regression equation to predict shipping time?

Answer 1

a) Scatter plot:

b-1)

Ʃx =	14524
Ʃy =	184
Ʃxy =	134691
Ʃx² =	10679344
Ʃy² =	1964
Sample size, n =	20
SSxx = Ʃx² - (Ʃx)²/n = 10679344 - (14524)²/20 =	132015.2
SSyy = Ʃy² - (Ʃy)²/n = 1964 - (184)²/20 =	271.2
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 134691 - (14524)(184)/20 =	1070.2

x̅ = Ʃx/n = 14524/20 = 726.2

y̅ = Ʃy/n = 184/20 = 9.2

Standard deviation, sx = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(10679344-(14524)²/20)/(20-1)] = 83.356

Standard deviation, sy = √[(Ʃy² - (Ʃy)²/n)/(n-1)] = √[(1964-(184)²/20)/(20-1)] = 3.778

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 1070.2/√(132015.2*271.2) = 0.179

b-2)

df = n-2 = 18

Critical value, t_c = T.INV(0.05, 18) = 1.734

b-3)

Test statistic :  

t = r*√(n-2)/√(1-r²) = 0.179 *√(20 - 2)/√(1 - 0.179²) = 0.771

b-4)

We cannot conclude that there is a positive correlation between distance and time.

c-1)

Coefficient of determination, r² = (SSxy)²/(SSxx*SSyy) = (1070.2)²/(132015.2*271.2) = 0.032

c-2)

3.2% variation in y is explained by the least squares model.

d)

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 271.2 - (1070.2)²/132015.2 = 262.5242714

Standard error, se = √(SSE/(n-2)) = √(262.52427/(20-2)) = 3.819

e)

No, we would not recommend using the regression equation to predict shipping time.