Income Frequency (1000s)

Under $10,000 4,187 Probability that family is selected

$10,000 - $14,999 3,653
$15,000 - $24,999 8,639 a. between $50,000 and $74,999, inclusive
$25,000 - $34,999 8,996 b. between $25,000 and $74,999, inclusive
$35,000 - $49,999 12,192 c. under $15,000
$50,000 - $74,999 15,676
$75,000 & over 18,192
Total 71,535
a. P(a)=Selected Random sample/ Sa

0.219137

b P(b)=Selected Random sample/ Sa

0.515328

c P(b)=Selected Random sample/ Sa

0.109597
ily is selected

and $74,999, inclusive

and $74,999, inclusive

cted Random sample/ Sample space

cted Random sample/ Sample space

cted Random sample/ Sample space

Military radar and missile detection systems are designed to warn a country against enemy attacks. A
reliability question is whether a detection system will be able to identify an attack and issue a warning.
Assume that a particular detection system has a 0.90 probability of detecting a missile attack. Answer the
following questions. (a) What is the probability that a single detection system will detect an attack? (b) If two
detection systems are installed in the same area and operate independently, what is the probability that at
least one of the systems will detect the attack? (c) If three systems are installed, what is the probability that
at least one of the systems will detect the attack
a. 0.9 Probability of detecting the attack is given
b. Let A and B be two systems
P(A)= 0.9 Missile is detected
P(A')= 0.1 Missile undetected

P(B) 0.9 Missile is detected

P(B') 0.1 Missile undetected

Probability of atleast 1 detects

P(A)*P(B)+P(A')*P(B)+P(A)*P(B')
0.99

c. A,B,C are three systems

P(A)= 0.9 Missile is detected

P(A')= 0.1 Missile undetected

P(B) 0.9 Missile is detected

P(B') 0.1 Missile undetected

P(C) 0.9 Missile is detected

P(C') 0.1 Missile undetected

1-P(A')P(B')P(C')
0.999
 Moderate Growth Low Growth
High Eco growth
0.3 0.5 0.2

will appreciate
will not

0.7 0.3 0.4 0.6 0.2

Given Rupee is appreciating, To find period is of high growth

using Bayes theorem

P= 0.4375
Low Growth

0.8
 Standard
Average
Deviatio
MPG
n
Automobile A 42 4
Automobile B 38 7 a To find which car can satisfy the decision making c

Assuming 95% confidence interval

For Car A

Mu 45

xbar(sample) 42
Std Dev 4

Z(calculated) -0.75
P(>45) p value 0.226627

But since probability of getting Mileage > 45 is high

b To find which car can satisfy the decision making c

Assuming 95% confidence interval

For Car A

Mu 39
xbar(sample) 42
Std Dev 4

Z(calculated) 0.75
P(<39) p value 0.226627

Since P value of Car B is more for Mileage being les

r can satisfy the decision making condition of Mileage being > 45 (to accept)

onfidence interval

For Car B

Mu 45

xbar(sample) 38
Std Dev 7

Z(calculated) -1
p value 0.158655

ility of getting Mileage > 45 is higher for car A we select Car A

r can satisfy the decision making condition of Mileage being < 39 (to reject)

onfidence interval

For Car B

Mu 39
xbar(sample) 38
Std Dev 7

Z(calculated) -0.142857
p value 0.556798

Car B is more for Mileage being less than 39 we will Choose Car A
 A
K boat L boat Using Ftest two sample test for variances
12 11.8 Null Hypothesis
13.1 12.1 Alternate hypothesis
11.8 12
12.6 11.6 Variance for K Boat sapmle data
14 11.8 Variance for L boat sample data
11.8 12
12.7 11.9 F Stat= var1/var2
13.5 12.6 F critical
12.4 11.4
12.2 12 Result
11.6 12.2 Conclusion
12.9 11.7

DOF=11 DOF=11 B Testing for performace of two boats

t-Test: Two-Sample Assuming Equal Variances

e test for variances
S1 sqaure = s2 square
S1 sqaure != s2 square

0.539090909090909
0.0947727272727272

5.68824940047962
0.354870359883879

Since, F(stat)> F -(Critical), We reject H0

We cannot assume equal variance for 2 population of boats

e of two boats

Ho: µ1-µ2=0 Null Hypothesis

suming Equal Variances H1: µ1-µ2≠0 (Two Tail)
α=5%

K boat L boat
Mean 12.55 11.925
Variance 0.539090909 0.094772727
Observations 12 12
Pooled Variance 0.316931818
Hypothesized Mean Difference 0
df 22
t Stat 2.719397672
P(T<=t) one-tail 0.00626043
t Critical one-tail 1.717144374
P(T<=t) two-tail 0.01252086
t Critical two-tail 2.073873068

P( two tail) 0.01252086

is lower than Aplha

Hence H0 is rejected
Hence two boats do not perform equally well

Average of L 11.925
Averag of K 12.55

Avg L< Avg K

K Boat is faster on an average compared to L boat
Null Hypothesis
 Stores for distinct
Observat
Country Consumers
ions
(Adult/Kid) H0:
1 USA 1 H1:
2 USA 1
3 USA 1
4 USA 2
5 USA 2
6 USA 1
7 USA 1
8 UK 2
9 UK 2
10 UK 1 Observed Frequency
11 UK 1
12 UK 2
13 UK 2
14 UK 2
15 UK 1
16 UK 1
17 Canada 1
18 Canada 2 Expected Frequency
19 Canada 1
20 Canada 2
21 Canada 1
22 Canada 1

Chi Sq P value
Fashion stores opening the kids stores is independent of Country
Fashion stores opening the kids stores is dependentof Country
Adult Kid
Count of Stores for distinct Consumers (Adult/Kid) Stores for distinct Consumers (Adult/Kid)
Country 1 2
Canada 4 2
UK 4 5
USA 5 2
Total Result 13 9

Row Labels Adult Kid Grand Total

Canada 4 2 6
UK 4 5 9
USA 5 2 7
Grand Total 13 9 22

Row Labels Adult Kid Grand Total

Canada 3.545455 2.45454545454545 6
UK 5.318182 3.68181818181818 9
USA 4.136364 2.86363636363636 7
Grand Total 13 9 22

Using Chi Sq test of independence

0.66478657 aplha 0.01

P value >0.01
Accept the Hypothesis

Hence ,
Fashion stores opening the kids stores is independent of Country
nct Consumers (Adult/Kid)
Total Result
6
9
7
22
 Rep 1 Rep 2 Rep 3 Rep 4
Dist 1 1 3 10 12
Dist 2 17 12 16 14
Dist 3 17 21 22 25
Dist 4 20 10 17 23
Dist 5 22 21 37 32
1 Using ANNOVA Test for checking hypoth
H0:
H1

Dist 1
Rep 1 1
Rep 2 3
Rep 3 10
Rep 4 12

using annova single factor

Anova: Single Factor

SUMMARY
Groups Count
Dist 1 4
Dist 2 4
Dist 3 4
Dist 4 4
Dist 5 4

ANOVA
Source of VariationSS
Between G 1011.3
Within Gro 407.5

Total 1418.8

P < 0.05
Hence Sales is dependent on

H0:
2 H1
Using ANNOVA Single factor
 Rep 1
1
17
17
20
22

Anova: Single Factor

SUMMARY
Groups
Rep 1
Rep 2
Rep 3
Rep 4

ANOVA
Source of Variation
Between G
Within Gro

Total

P value > 0.05

hence we accept the Null hyp

Sales may be indendent of Re

Or Representatives are not s
NOVA Test for checking hypothesis
Sales is indendent of District
Sales is dependednt on District

Dist 2 Dist 3 Dist 4 Dist 5

17 17 20 22
12 21 10 21
16 22 17 37
14 25 23 32

using annova single factor test

Sum Average Variance

26 6.5 28.33333
59 14.75 4.916667
85 21.25 10.91667
70 17.5 31
112 28 60.66667

df MS F P-value F crit
4 252.825 9.306442 0.000548 3.055568
15 27.16667

19

We reject the Nul Hypothesis

Hence Sales is dependent on District

Sales is indendent of Representatives

Sales is dependednt on Representatives
Using ANNOVA Single factor
 Rep 2 Rep 3 Rep 4
3 10 12
12 16 14
21 22 25
10 17 23
21 37 32

Anova: Single Factor

SUMMARY
Count Sum Average Variance
5 77 15.4 69.3
5 67 13.4 59.3
5 102 20.4 104.3
5 106 21.2 67.7

SS df MS F P-value F crit
216.4 3 72.13333 0.959858 0.435593 3.238872
1202.4 16 75.15

1418.8 19

P value > 0.05

hence we accept the Null hypothesis

Sales may be indendent of Representatives

Or Representatives are not strong factor to determine sales
 Technolog
Trained
Food Entertainm y
Event Planners Overall workfor
Services ent applicatio
ce
Eventena Events 94.4 94.6 90.9 97.8 ns 91
Crystal Events 93 96.7 84.2 96.7 99
way4events - Event 92.9 100 100 88.5 90
Lolipop events 91.3 88.6 94.8 97.1 88
Tina events 90.5 95.1 87.9 91.2 91
Carribean nights 90.3 92.5 82.1 98.8 91
Charlie events 90.2 96 86.3 92 99
Il Fornaio 89.9 92.6 92.6 88.9 87
Joe's dreams 89.4 94.7 85.9 90.8 91
Johnny Carino 89.2 90.6 83.3 90.5 99
Lone Star 89.2 90.9 82 88.6 90
LongHorn 89.1 93.1 93.1 89.7 88
Maggiano events 88.7 92.6 78.3 91.3 91
McGrath events 87.2 93.1 91.7 73.6 91
Olive Garden 87.2 91 75 89.7 88
Outback Steakhouse 86.6 94.4 78.1 91.6 91
IMC events 86.2 95.5 77.4 90.9 91
Grace events 86.1 94.9 76.5 91.5 88
1 SUMMARY OUTPUT OF FOOD SERVICE WITH OVERALL

Regression Statistics
Multiple R 0.254503
R Square 0.064772
Adjusted R Square 0.00632
Standard Error 2.366007
Observations 18

ANOVA
df SS MS
Regression 1 6.203281 6.203281
Residual 16 89.56783 5.597989
Total 17 95.77111

Coefficients
Standard Error t Stat
Intercept 68.05696 20.39876 3.336328
Food Services 0.229044 0.217583 1.052676

2 SUMMARY OUTPUT OF ENTERTAINMENT WITH OVERALL

Regression Statistics
Multiple R 0.636679
R Square 0.405361
Adjusted R Square 0.368196
Standard Error 1.886619
Observations 18

ANOVA
df SS MS
Regression 1 38.82184 38.82184
Residual 16 56.94927 3.559329
Total 17 95.77111

Coefficients
Standard Error t Stat
Intercept 71.47724 5.481962 13.03862
Entertainment 0.210902 0.06386 3.302585
3 SUMMARY OUTPUT OF TRAINED WORKFORCE WITH OVERALL

Regression Statistics
Multiple R 0.470505
R Square 0.221375
Adjusted R Square 0.172711
Standard Error 2.158845
Observations 18

ANOVA
df SS MS
Regression 1 21.20135 21.20135
Residual 16 74.56976 4.66061
Total 17 95.77111

Coefficients
Standard Error t Stat
Intercept 70.79418 8.795489 8.04892
Trained workforce 0.205652 0.096421 2.13285

4 SUMMARY OUTPUT OF TECHNOLOGY APPLICATION WITH OVERALL

Regression Statistics
Multiple R 0.256869
R Square 0.065982
Adjusted R Square 0.007606
Standard Error 2.364476
Observations 18

ANOVA
df SS MS
Regression 1 6.319144 6.319144
Residual 16 89.45197 5.590748
Total 17 95.77111

Coefficients
Standard Error t Stat
Intercept 74.82404 13.83636 5.407783
Technology applications 0.160929 0.15137 1.063149

5 SUMMARY OUTPUT WITH ALL INDEPENDENT VARIABLES AGAINST OVERALL

Regression Statistics
Multiple R 0.894929
R Square 0.800897
Adjusted R Square 0.739635
Standard Error 1.211112
Observations 18

ANOVA
df SS MS
Regression 4 76.70281 19.1757
Residual 13 19.0683 1.466793
Total 17 95.77111

Coefficients
Standard Error t Stat
Intercept 27.31309 12.6589 2.157619
Food Services 0.061906 0.117641 0.526229
Entertainment 0.241973 0.042663 5.671696
Trained workforce 0.236697 0.055152 4.291698
Technology applications 0.154914 0.081652 1.897237
 Food Services
102
100
98
96
F Significance F 94 f(x) = 0.282792305728936 x + 68.4004710304662

Axis Title
92 R² = 0.0647719419862043
1.108127 0.308129 90
88
86
84
82
85 86 87 88 89 90 91 92 93 94 9
P-value Lower 95%Upper 95%Lower 95.0%
Upper 95.0% Axis Title
0.004186 24.81352 111.3004 24.81352 111.3004
0.308129 -0.232211 0.690299 -0.232211 0.690299 Food Services Linear (Food Services)

Trained workforce
120

100
f(x) = 1.07645543773348 x − 5.30001624242966
80 R² = 0.221375183045555
Axis Title

60

40

20

0
85 86 87 88 89 90 91 92 93 94 9
F Significance F Axis Title
10.90707 0.004495
Trained workforce Linear (Trained workforce)

P-value Lower 95%Upper 95%Lower 95.0%

Upper 95.0%
6.125E-10 59.856 83.09848 59.856 83.09848
0.004495 0.075525 0.346278 0.075525 0.346278
 F Significance F
4.54905 0.048774

P-value Lower 95%Upper 95%Lower 95.0%

Upper 95.0%
5.123E-07 52.14858 89.43979 52.14858 89.43979
0.048774 0.001248 0.410056 0.001248 0.410056

F Significance F
1.130286 0.303498

P-value Lower 95%Upper 95%Lower 95.0%

Upper 95.0%
5.806E-05 45.49227 104.1558 45.49227 104.1558
0.303498 -0.159961 0.481819 -0.159961 0.481819
 F Significance F
13.07322 0.000173

P-value Lower 95%Upper 95%Lower 95.0%

Upper 95.0%
0.050252 -0.034803 54.66098 -0.034803 54.66098
0.60759 -0.192243 0.316055 -0.192243 0.316055
7.647E-05 0.149805 0.334142 0.149805 0.334142
0.000877 0.117548 0.355846 0.117548 0.355846
0.080231 -0.021485 0.331313 -0.021485 0.331313
ices Entertainment
120

100
f(x) = 1.92203633663596 x − 86.5038529363992
80 R² = 0.40536065589185
x + 68.4004710304662
Axis Title

60

40

20

0
0 91 92 93 94 95 85 86 87 88 89 90 91 92 93 94 95
tle Axis Title

near (Food Services) Entertainment Linear (Entertainment)

kforce Technology applications

100
98
96
− 5.30001624242966 94
92 f(x) = 0.410005336798385 x + 54.6287444601712
Axis Title

90 R² = 0.065981733162363
88
86
84
82
80
91 92 93 94 95 85 86 87 88 89 90 91 92 93 94 95
tle Axis Title

ear (Trained workforce) Technology applications Linear (Technology applications)