Question

1. From Statistics and Data Analysis from Elementary to Intermediate by Tamhane and
Dunlop, pg 339. The following table gives the eye color and hair color of 592 students.

Eye Hair Color Row
Color Black Brown Red Blond Total
Brown 68 119 26 7 220
Blue 20 84 17 94 215
Hazel 15 54 14 10 93
Green 5 29 14 16 64
Column Total 108 286 71 127 592

a) What test should we use to test that the eye color and hair color are associated? Give the null
and alternative hypothesis.
b) Conduct the test at α = 0.05. What do you conclude? Give more than reject or fail to reject
H0.

2. The following data is looking at how long it takes to get to work. Let x = commuting
distance (miles) and y = commuting time (minutes)

x 15 16 17 18 19 20
y 42 35 45 42 49 46

a) Give a scatterplot of this data and comment on the direction, form, and strength of this relationship.

b) Determine the least-squares estimate equation for this data set.
c) Give the coefficient of determination, R2, comment on what that means.
d) Give the residual plot based on the least-squares estimate equation.
e) Test if this least-squares estimate equation specify a useful relationship between commuting
distance and commuting time.

Answer 1

1.

a) What test should we use to test that the eye color and hair color are associated? Give the null
and alternative hypothesis..

We shall be using a test for independence/association of attributes between eye color and the hair color.

Null Hypothesis: There is no association between eye color and the hair color

Alternate Hypothesis: There is an association between eye color and hair color.

b). Conduct the test at α = 0.05. What do you conclude? Give more than reject or fail to reject
H0.

Level of significance

Test statistic:

where is the observed frequency in the   cell, is the expected frequency in the cell and is given by   where is the total of row, is the total of column and T is the total of all the values. We shall first calculate the expected frequency first and then calculate the Chi-square.

The observed frequencies:

OBS	Black	Brown	Red	Blond	Total
Brown	68	119	26	7	220
Blue	20	84	17	94	215
Hazel	15	54	14	10	93
Green	5	29	14	16	64
Total	108	286	71	127	592

The expected frequencies.

EXPECTED	Black	Brown	Red	Blond	Total
Brown	40.13514	106.2838	26.38514	47.19595	220
Blue	39.22297	103.8682	25.78547	46.12331	215
Hazel	16.96622	44.92905	11.15372	19.95101	93
Green	11.67568	30.91892	7.675676	13.72973	64
Total	108	286	71	127	592

CHiSQ	Black	Brown	Red	Blond	Total
Brown	19.34591	1.521419	0.005622	34.23417	55.10712
Blue	9.421078	3.80046	2.993334	49.69672	65.91159
Hazel	0.227865	1.831378	0.726335	4.96329	7.748867
Green	3.816879	0.119094	5.210887	0.375399	9.522259
Total	32.81173	7.27235	8.936177	89.26958	138.2898

The critical value of at 9 df is 16.9190.

The calculated value of is 138.2898>the critical value and so we reject the null hypothesis. Hence, we conclude that the eye color and the hair color is associated.

2. The given data:

x	y
15	42
16	35
17	45
18	42
19	49
20	46

a). the scatter plot.

The scatter plot shows an upward trend of the points and the points seem to form a line but the relationship is weak since all the points are bit away from the line.

b) Determine the least-squares estimate equation for this data set.

For the calculation of the least square estimates:

	x	y	x^2	y^2	x*y
	15	42	225	1764	630
	16	35	256	1225	560
	17	45	289	2025	765
	18	42	324	1764	756
	19	49	361	2401	931
	20	46	400	2116	920
Total	105	259	1855	11295	4562

Therefore the estimated linear equation is

c) Give the coefficient of determination, R2, comment on what that means.

The coefficient of determination is . The correlation is given by

  

  

The coefficient of determination is This implies that 43.31% of the variation of the data is being explained by this model.

d) Give the residual plot based on the least-squares estimate equation.

e). Test if this least-squares estimate equation specify a useful relationship between commuting
distance and commuting time.

For this, we shall create the ANOVA table. We need to calculate the F-value for the Regression SS.

The Total sum of squares is

The regression sum of squares is

The error sum of squares ESS=TSS-RSS=114.83333-49.7285=65.1052

We shall form the ANOVA table:

	df	SS	MS	F	Significance F
Regression	1	49.72857	49.72857	3.055295	0.15539
Residual	4	65.10476	16.27619
Total	5	114.8333

From the Table above, we can see that the p-value is 0.1554>0.05. Hence we fail to reject the null hypothesis and conclude that there is not enough evidence to that teh relationship between X and Y is linear. There fore,the equation does not specify a useful relationship between commuting distance and commuting time.