Quantitative Analysis

This document discusses the concept of racial steering in housing and analyzes data to determine if there is evidence of racial steering. It also includes a discussion on the survival rates of men and women on a sunken ship, the impact of the Internet on daily life, and the relationship between income level and cybershopping. Additionally, it examines the assessment methods used by companies to evaluate the success of their ERP systems and analyzes the percent change in real GDP by region in a certain country.

Running Head: QSO/501 1
Quantitative Analysis
By (Name of Student)
(Institutional Affiliation)
(Date of Submission)

QSO/510
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Question 1
A subtle form of racial discrimination in housing is "racial steering." Racial steering
occurs when real estate agents show prospective buyers only homes in neighborhoods
already dominated by that family's race. Tenants at a large apartment complex recently
filed a lawsuit alleging racial steering. The plaintiffs claimed that potential renters were
steered to Section A, while African-American renters were steered to Section B. The
accompanying table displays the data that were presented in court to show the locations of
recently rented apartments. Do you think there is evidence of racial steering?
Data to use on this question
- White
African-
American Total
Section 91 9 100
Section 91 28 119
Total 192 37 219
Solution
To solve this problem, we need to first state the null and alternative hypotheses under
consideration before we compute the χ2 statistic.
Ho: There is no association between race and the section of the apartment complex.
HA: There is an association between race and the section of the apartment complex.
The formula for computing χ2 statistic is given by;
χ2∗=∑ (Oi−Ei) 2/Ei, where Oi and Ei is the ith observation and the ith expected count.

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From the data given, we then calculate the expected count using the formula;
E= (row total × column total)/sample size.
We compute the χ2 statistic as follows;
Observed (O) Expected
(
Observed-Expected)^2/Exp
ected
Section A, white 91 78.38776 0.557762051
Section A,African
American 9 13.61224 3.211941445
Section B, white 91 88.61224 0.49340495
Section B,African
American 28 15.38776 2.84133089
TOTAL 219 196 7.104439336
χ2 = 7.104439336,
χ2 = 7.104 (rounded to 3 decimal .places)
Find the P-value.
Solution
Degrees of freedom = (2-1)*(2-1) = 1, χ2 = 7.1044
P-value from chi-square calculator is 0.007689
Decision

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State your conclusion at the significance level α = 0.05.
Solution
If the p- value less than or equal to the alpha α or level of significance, then we reject the null
hypothesis in favor of the alternative hypothesis, otherwise we fail to reject the null hypothesis.
In this case, the p-value, 0.007689, is less than alpha (α = 0.05). That is 0.007689<0.05.
Therefore, we reject the null hypothesis.
Reject H0. There is sufficient evidence to conclude that there is an association between race and
the section of the apartment complex in which people live.
Question 2
Newspaper headlines at the time and traditional wisdom in the succeeding decades have
held that women and children escaped a sunken ship in greater proportion than men.
Here's a table with the relevant data. Do you think that survival was independent of
whether the person was male or female? Defend your conclusion.
Female Male Total
Alive 393 315 708
Dead 116 1348 1464
Total 509 1663 2172
Solution
A. H0: Survival was independent of gender on the ship. 0
HA: There is an association between survival and gender on the ship.
B. Are the conditions for inference satisfied?

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Answer
B. Yes, because the data are counts, the data in groups are independent, and the groups have at
least 5 observations.
Find the χ2 statistic
Observed Expected
(Observed-Expected)^2/
Expected
393 152.4106 328.0979
315 471.7594 99.4903
116 345.7594 143.5152
1348 1140.2406 43.5187
TOTAL 2172 2136.0000 614.6221
χ2 statistic = 614.62 (to 2 decimal places).
Find the P-value.
Solution
Degrees of freedom = (2-1)* (2-1) = 1, χ2 = 614.6221
P-value from chi-square calculator is < 0.00001
P- Value is 0.000 (to 2 decimal places)
Interpret the p-value. Choose the correct answer below.
P- Value is less than 0.025. So, we reject the null hypothesis and conclude that there is evidence
of a significant difference between the proportion of males and females who survived at the
0.025 level of significance.

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C. We reject the null hypothesis and conclude there is strong evidence of an association between
survival and gender on the ship. Females were more likely to survive.
Question 3.
A polling institute routinely conducts surveys to gauge the impact of the Internet and
technology on daily life. A recent survey asked respondents if they read online journals or
blogs, an Internet activity of potential interest to many businesses. A subset of the data
from this survey shows responses to this question. Test whether reading online journals or
blogs is independent of generation. Use a significance level of α = 0.05.
Age_Group
Yesterda
y
Yes but
Not_Yesterday No
Gen-Y(18-30) 32 37 69
Gen-X(31-42) 11 40 131
Trailing_Boomers(43-
52) 15 34 142
Leading_Boomers(53-
61) 9 22 81
Matures(62+) 6 25 102
Solution
We first state the respective null and alternative hypotheses and make the decision based on the
finding.
Answer; A

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Ho: Generation and online journal or blog reading are independent.
HA: Generation and online journal or blog reading are not independent.
We compute the χ2 statistic as shown in the table below;
Obs
erve
d
Exp
ecte
d
(Observed-
Expected)^
2/Expected
Obs
erve
d
Exp
ecte
d
(Obse
rved-
Expec
ted )^
2/Exp
ected
32
12.6
831 16.16116 142
121.
637
8
0.1564
38
37
25.5
353 4.288571 9
11.8
842
0.6999
72
69
88.7
816 6.917286 22
23.9
268
0.1551
63
11
19.3
742 0.987583 81
83.1
891
0.2782
19
40
39.0
067 0.231761 6
13.6
818
1.6020
74
131
135.
619
2 0.401685 25
27.5
459
0.4564
53

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15
17.3
768 0.325099 102
95.7
723
0.7068
33
34
34.9
854 0.11267
χ2 statistic = 33.48096835
= 33.48 (2 decimal places)
The degrees of freedom = (5-1)*(3-1) = 8, χ2 = 33.48097
P-value from chi-square calculator is 5E-05= 5*10 ^ (-5) = 0.00005
P- Value = 0.000 (3 decimal places)
Conclusion based on the findings;
Since p-value is less than alpha or the significance level (0.002454< 0.05), we reject the null
hypothesis. Thus;
The P-value is less than the significance level. There is sufficient evidence to reject the null
hypothesis.
Hence, the test concludes that reading online journals or blogs not independent of generational
age
Question 4.
It has become more common for shoppers to "comparison shop" using the Internet.
Respondents to a survey in 2013 who owned cell phones were asked whether they had, in
the past 30 days, looked up the price of a product while they 3: More Info were in a store to
see if they could get a better price somewhere else.

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_ <$30K
$30K-
$49.9K
$50K-
$74.9K >$75K
Yes 238 113 134 182
No 678 360 264 426
Solution
We first, state the null and alternative hypotheses for testing whether income level and
cybershopping are related. The null and alternative hypotheses are as follows.
H0: Income level and cybershopping are independent.
HA: Income level and cybershopping are not independent.
Then we compute the χ2 statistic as follows;
Observe
d Expected
Observed
-Expected
(Observed-
Expected)^2/
Expected
238 214.02993 -37.02993 6.406654
113 126.29577 -33.29577 8.777874
134 103.77993 62.22007 37.30333
182 143.89437 8.10563 0.456594
678 612.97007 37.02993 2.237003
360 361.70423 33.29577 3.064958
264 297.22007 -62.22007 13.02515
426 412.10563 -8.10563 0.159428

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TOTAL 2272.00000 71.431
χ2 statistic = 71.43(2 decimal places)
Degrees of freedom = (4-1)*(2-1) = 3, χ2 = 71.431
P-value from chi-square calculator is < 0.00001
P- Value is 0.000 (3 decimal places)
Choose alpha = 0.05, then P is less than 0.05, we reject the null hypothesis.
State the appropriate conclusions about the hypotheses. Choose the correct answer below.
Answer
B. Reject the null hypothesis. Income level and cybershopping are independent.
We then compute and examine the standardized residuals. After examining the pattern, if any, do
they show that would be of interest to retailers concerned about cybershopping comparisons?
Standardized residuals = residuals / standard deviations of the residuals
standardized residuals
<30K 30K- 49.9k 50K-74.99K >75k
YE
S 0.86482 0.77761 1.45313 0.18930
NO 0.86482 0.77761 1.45313 0.18930
The pattern shown therefore indicates that shoppers are less likely to cybershop for comparison
prices if they earn more money.

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Question 5
In a recent study of enterprise resource planning (ERP) system effectiveness, researchers
asked companies how they assessed the success of their ERP systems. Out of 420
manufacturing companies surveyed, they found that 214 used return on investment (ROI),
160 used reductions in inventory levels, 33 used improved data quality, and 13 used on-
time delivery. In a survey of 440 service firms, 230 used ROI, 160 used inventory levels, 30
used improved data quality, and 20used on-time delivery.
Solution
Stating the null and alternative hypotheses
H0: The proportions of assessment method are the same for each industry sector.
HA: The proportion of assessment method is different for at least one industry sector.
Computing the χ2 statistic.
Observed Expected
(Observed-Expected)^2/
Expected
214 139.2 13.15977
160 120 3.333333
33 91.2 26.54211
13 24.6 7.518699
230 92.8 19.73966
160 80 5

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30 60.8 39.81316
20 16.4 11.27805
TOTAL 860 625 126.3848
χ2 statistic = 126.385 (to 3 decimal places)
Find the p- value
P –value is 0.0000
State your conclusion at the significance level α = 0.05. Choose the correct answer below.
The P – Value is less than the significance level α = 0.05; we reject null hypothesis in favor of
the alternative hypothesis.
A. Reject H0. There is sufficient evidence that the proportion of assessment method is
different for at least one industry sector.
Question 6
A government's economic analysis branch recently released figures that they claimed
showed a growth spurt in the western region of a certain country that is divided into 51
states. Using the accompanying table, where the cell values are numbers of states,
determine if the percent change in real GDP by state was independent of region of the
country. Use α = 0.05
Region T
o
p
_
Bo
tto
m
_6

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4
0
%
0
%
West
1
3 2
Midwest 3 9
Southeast 3 9
Northeast 2 10
Answer B
Stating the null and
alternative hypothesis;
H0: GDP % change is
independent of region.
HA: GDP % change is
not independent of
region.
What is the test
statistic?
Observed

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TOTAL
χ2 = 13.91( 2 decimal
places)
What is the P-value?
Degrees of freedom =
3, χ2 = 13.912
P-value from chi-
square calculator is
P-value = 0.003027
P-value = 0.003( 3
decimal places)

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What is the proper
conclusion?
P is less than 0.05, so
we reject the null
hypothesis in favor of
the alternative
hypothesis
Reject H0. There is
sufficient evidence to
conclude that GDP %
change is not
independent of region.
References
Field, A. P., & Gillett, R. (2010). How to do a meta‐analysis. British Journal of Mathematical
and Statistical Psychology, 63(3), 665-694.
McHugh, M. L. (2013). The chi-square test of independence. Biochemia medica: Biochemia
medica, 23(2), 143-149.

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Satorra, A., & Bentler, P. M. (2010). Ensuring positiveness of the scaled difference chi-square
test statistic. Psychometrika, 75(2), 243-248.
Sullivan, G. M., & Feinn, R. (2012). Using effect size—or why the P value is not
enough. Journal of graduate medical education, 4(3), 279-282.