Question

I know this is a long problem but I couldn't break it up because you need all the information. But it is only counted as one question on my homework.

Fuming because you are stuck in traffic? Roadway congestion is a costly item, both in time wasted and fuel wasted. Let x represent the average annual hours per person spent in traffic delays and let y represent the average annual gallons of fuel wasted per person in traffic delays. A random sample of eight cities showed the following data.

x (hr)	29	5	18	37	22	25	15	5
y (gal)	48	3	32	53	31	38	26	9

Verify that Σx = 156, Σx² = 3918, Σy = 240, Σy² = 9308, and Σxy = 6011.

Compute r._____________?

The data in part (a) represent average annual hours lost per person and average annual gallons of fuel wasted per person in traffic delays. Suppose that instead of using average data for different cities, you selected one person at random from each city and measured the annual number of hours lost x for that person and the annual gallons of fuel wasted y for the same person.

x (hr)	24	4	20	40	19	25	2	38
y (gal)	62	8	14	51	23	35	4	71

(b) Compute x and y for both sets of data pairs and compare the averages.

	x	y
Data 1	?	?
Data 2	?	?

Compute the sample standard deviations s_x and s_y for both sets of data pairs and compare the standard deviations.

	sx	sy
Data 1	?	?
Data 2	?	?

Verify that Σx = 172, Σx² = 5026, Σy = 268, Σy² = 13,516, and Σxy = 7858.

Compute r.__________?

List some reasons why you think hours lost per individual and fuel wasted per individual might vary more than the same quantities averaged over all the people in a city.

Answer 1

X	Y	XY	X²	Y²
29	48	1392	841	2304
5	3	15	25	9
18	32	576	324	1024
37	53	1961	1369	2809
22	31	682	484	961
25	38	950	625	1444
15	26	390	225	676
5	9	45	25	81

Ʃx =	156
Ʃy =	240
Ʃxy =	6011
Ʃx² =	3918
Ʃy² =	9308
Sample size, n =	8
SSxx = Ʃx² - (Ʃx)²/n = 3918 - (156)²/8 =	876
SSyy = Ʃy² - (Ʃy)²/n = 9308 - (240)²/8 =	2108
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 6011 - (156)(240)/8 =	1331

a)

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 1331/√(876*2108) = 0.979

----

For Data Set 1:

x̅ = Ʃx/n = 156/8 = 19.5  

y̅ = Ʃy/n = 240/8 = 30  

Standard deviation, sx = √[(Ʃx² - (Ʃx)²/n)/(n-1)] =    √[(3918-(156)²/8)/(8-1)] =    11.1867

Standard deviation, sy = √[(Ʃy² - (Ʃy)²/n)/(n-1)] =    √[(9308-(240)²/8)/(8-1)] =    17.3535

-----------

X	Y	XY	X²	Y²
24	62	1488	576	3844
4	8	32	16	64
20	14	280	400	196
40	51	2040	1600	2601
19	23	437	361	529
25	35	875	625	1225
2	4	8	4	16
38	71	2698	1444	5041

Ʃx =	172
Ʃy =	268
Ʃxy =	7858
Ʃx² =	5026
Ʃy² =	13516
Sample size, n =	8
SSxx = Ʃx² - (Ʃx)²/n = 5026 - (172)²/8 =	1328
SSyy = Ʃy² - (Ʃy)²/n = 13516 - (268)²/8 =	4538
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 7858 - (172)(268)/8 =	2096

For data Set 2 :

x̅ = Ʃx/n = 172/8 = 21.5  

y̅ = Ʃy/n = 268/8 = 33.5  

Standard deviation, sx = √[(Ʃx² - (Ʃx)²/n)/(n-1)] =    √[(5026-(172)²/8)/(8-1)] = 13.7737

Standard deviation, sy = √[(Ʃy² - (Ʃy)²/n)/(n-1)] =    √[(13516-(268)²/8)/(8-1)] = 25.4615

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 2096/√(1328*4538) = 0.854