Ladies Professional Golf Association
Winnings
(LPGA)
VS
Senior Professional Golf Association
Winnings
(SPGA)
2-Sample t-Test
TOUR
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
LPGA
WINNINGS
1591959
1337253
956926
863816
757844
679929
663356
584246
583796
577875
572940
538054
512273
501798
497640
484759
447903
440498
410973
405142
370162
369176
367258
355989
354131
322308
303929
301086
297973
296347
TOUR
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
Senior
EARNINGS
2515705
2025232
1911640
1513524
1493282
1327658
1167176
1118377
1108245
1087284
1051357
1039334
997318
993291
988778
951072
882532
869839
857746
816342
754046
743841
737860
726674
715035
710749
683314
638621
635095
631046
MIN
Q1
MED
Q3
MAX
LPGA
296347
367258
491200
584246
1591959
SPGA
631046
737860
969925
1118377
2515705
MEAN
SD
LPGA
558245
299096
SPGA
1056400
446668
We can see from the graph and the relationships between the
mean and median for each data set (mean>median), that both
data sets are skewed right. SPGA has a larger center than
LPGA. This can be seen from the graphs an comparing the
median and mean for each data set. SPGA has a larger
spread than LPGA. Both data set have outliers on the high
end of the data.
H o : L  S (Average LPGA Winning equals average SPGA winnings)
H o : L  S (Average LPGA Winning equals average SPGA winnings)
Even Though Both graphs are skewed right combined sample
size is equal to 60. This was not a random sample since I
gathered the top 30 for each tour so this might cause a problem
with conclusion.
x L  558, 245
xS  1, 056, 400
x L  xS  558, 245  1, 056, 400  498,155
t  5.0757
p  0.0000278
df  50.653
I will Reject Ho because p < 0.05 and | t | >1.675. My
data is statistically significant and I am able to conclude
that average winnings on LPGA tour are less than
average winning for SPGA tour.
Since this was a Reject Ho conclusion it is possible that
we have committed a Type I Error which would be
concluding that the average LPGA winnings is less than
SPGA, when in reality the average LPGA winnings are
not less than SPGA.
H o : L  S (Average LPGA Winning equals average SPGA winnings)
H o : L  S (Average LPGA Winning equals average SPGA winnings)
The 90% confidence interval for the difference in average
winnings(LPGA - SPGA) is (-66000,-33000). Since all the
values are negative we can conclude that average
winning for LPGA is less than average wining for SPGA
Husband’s Age
VS
Wife’s Ages
Matched Pair t-Test
Husband
22
38
31
42
23
55
24
41
26
24
19
42
34
31
45
33
54
20
43
24
40
26
29
32
36
68
19
52
24
22
29
54
35
22
Wife
21
42
35
24
21
53
23
40
24
23
19
38
32
36
38
27
47
18
39
23
46
25
27
39
35
52
16
39
22
23
30
44
36
21
Husband -Wife
1
-4
-4
18
2
2
1
1
2
1
0
4
2
-5
7
6
7
2
4
1
-6
1
2
-7
1
16
3
13
2
-1
-1
10
-1
1
Husband
44
33
21
31
21
35
23
51
38
30
36
50
24
27
22
29
36
22
32
51
28
66
20
29
25
54
31
23
25
27
24
62
35
26
Wife
44
37
20
23
22
42
22
47
33
27
27
55
21
34
20
28
34
26
32
39
24
53
21
26
20
51
33
21
25
25
24
60
22
27
Husband -Wife
0
-4
1
8
-1
-7
1
4
5
3
9
-5
3
-7
2
1
2
-4
0
12
4
13
-1
3
5
3
-2
2
0
2
0
2
13
-1
Husband
24
37
22
24
27
23
31
32
23
41
71
26
24
25
46
24
18
26
25
29
34
26
51
21
23
26
20
25
32
48
54
60
Wife
23
36
20
27
21
22
30
37
21
34
73
33
25
24
37
23
20
27
22
24
39
18
50
20
23
24
22
32
31
43
47
45
Husband -Wife
1
1
2
-3
6
1
1
-5
2
7
-2
-7
-1
1
9
1
-2
-1
3
5
-5
8
1
1
0
2
-2
-7
1
5
7
15
Husband-Wife
MIN
Q1
MED
Q3
MAX
-7
-1
1
4
18
MEAN
SD
1.92
5.04661
Husband-Wife
We can see from the graph and the relationships between the
mean and median for each data set (mean>median), that the
data sets is skewed right. We can also see that there are
outliers on the high end of the data set.
H o : H W  0(Average Husband's age and average Wife's age are equal)
H o : H W  0(Average Husband's age is greater than average Wife's age)
Even though the graph is slightly skewed right tah sample size
was 100. The data is a SRS
xH W  1.92
t  3.805
p  0.000123
df  99
I will Reject Ho because p < 0.05 and | t | >1.66. My data
is statistically significant and I am able to conclude that
average Husband’s age is greater than average
Husband’s are older than their Wife’s age
Since this was a Reject Ho conclusion it is possible that
we have committed a Type I Error which would be
concluding that the average Husband age is greater than
the average age of their Wife’s, when in reality the
average age of a Husband is not greater than the
average age of their Wife’s
H o : H W  0(Average Husband's age and average Wife's age are equal)
H o : H W  0(Average Husband's age is greater than average Wife's age)
The 90% confidence interval for the average difference in
ages(Husband-Wife) is (1.0821,2.7579). Since all the
values are positive we can conclude that average
Husband’s age is greater than the average Wife’s age
Football Injuries
VS
Baseball Injuries
2-Propotion z-Test
Sport
Injuries
Participants Proportion
Football
334420
20100000
0.01664
Baseball
326714
30400000
0.01075
Baseball
Football
1%
2%
98%
Injuries
Injuries
Participants
Participants
99%
H o : PF  PB (Equal Propotion of Football and Baseball Players get injured)
H o : PF  PB (A greater propootion Football players get injured)
It is reasonable to assume SRS
20100000(0.01664)>10
20100000(1-0.01664)>10
30400000(0.01075)>10
30400000(1-0.01075)>10
p̂F  0.01664
p̂B  0.01075
z  180.266
p0
p̂F  p̂B  0.01664  0.01074  0.0059
I will Reject Ho because p < 0.05 and z >1.645. My data
is statistically significant and I am able to conclude that a
higher proportion of Football players get injured
Since this was a Reject Ho conclusion it is possible that
we have committed a Type I Error which would be
concluding a higher proportion of Football players get
injured when in reality it is not true that a higher
proportion of Football players get injured
H o : PF  PB (Equal Propotion of Football and Baseball Players get injured)
H o : PF  PB (A greater propootion Football players get injured)
The 90% confidence interval for the difference in
proportion of injuries (Football - Baseball) is
(0.00583,0.00595). Since all the values are positive we
can conclude that proportion of Football players who get
injured is greater than the proportion of Baseball players
who get injured
Spending Money For Space
Exploration
VS
Political Perspective
Chi-Squared Test for Independence
Conservative
Liberal
Moderate
Just
Right
212
164
214
Too
Little
36
50
47
Too
Much
176
162
174
Moderate
250
214
212
200
176
164
174
162
150
Just Right
Too Little
40%
Too Much
100
49%
Just Right
Too Little
Too Much
50
50
36
47
11%
0
Conservative
Liberal
Moderate
Conservative
Liberal
42%
Just Right
50%
Too Little
43%
44%
Too Little
Too Much
8%
Just Right
Too Much
13%
Too Much
250
214
212
200
176
174
162
164
150
Conservative
34%
34%
Liberal
Conservative
Moderate
100
Liberal
Moderate
50
36
50 47
0
32%
Just Right
Too Little
Too Much
Just Right
Too Little
27%
36%
36%
28%
35%
Conservative
Conservative
Liberal
Liberal
Moderate
Moderate
38%
H o : There is no relationshipe between spening beliefs and political perspective
H o : There is a relationshipe between spening beliefs and political perspective
Conservative
Just
202.6
Right
Liberal
179.6
Moderate
207.8
Too
Little
45.7
40.5
46.8
Too
Much
175.8
155.9
180.3
It is reasonable to assume a SRS and all expected counts
are greater than 5
 2  6.724
p  0.15
df  4
2

I will Fail to Reject Ho because p > 0.05 and < 9.49.
My data is not statistically significant and I am unable to
conclude that there is a relationship between spending
perspective and political perspective
Since this was a Fail to Reject Ho conclusion it is possible
that we have committed a Type II Error which would be
concluding that there is no relationship between spending
perspective and political perspective when in fact a
relationship exist
Per Capita Income
VS
Governor's Salary
Linear Regression t-Test
PerCapitaInco me
20842
25305
22364
19585
26570
27051
36263
29022
25255
24061
26034
20478
28202
23604
23102
24379
20657
20680
22078
28969
31524
25560
26797
18272
24001
GovernorSalary
87643
81648
75000
60000
131000
70000
78000
107000
110962
115939
94780
75000
130261
77200
104352
85225
95526
95000
70000
120000
90000
127300
114506
83160
112755
PerCapitaInco me
20046
23803
26791
28047
32654
19587
30752
23345
20271
24661
20556
24393
26058
25760
20755
21447
23018
23656
20432
23401
26438
26718
18957
24475
22648
GovernorSalary
78246
65000
90000
90547
85000
90000
130000
107132
75372
115762
101140
88300
105035
69900
106078
82271
85000
115345
90700
105402
124855
121000
90000
115899
95000
H o :   0(No realtionship between Per Capita Income and Governer's Salary)
H o :   0(Positive realtionship between Per Capita Income and Governer's Salary)
Te scatter plot and r2 value indicate a weak linear fit. The
residual plot shows an unequal spread of the residual values.
These fact could result in incorrect results with test.
b  0.05778
t  2.101
p  0.02 df  48
I will Reject Ho because p < 0.05 and t >1.677. My data
is statistically significant and I am able to conclude there
is a linear relationship between Per Capita Income and
Governor's Salary
Since this was a Reject Ho conclusion it is possible that
we have committed a Type I Error which would be
concluding that there is a linear relationship between Per
Capita Income and Governor's Salary when no
relationship exist
H o :   0(No realtionship between Per Capita Income and Governer's Salary)
H o :   0(Positive realtionship between Per Capita Income and Governer's Salary)
The 90% confidence interval for the slope of the line of
best fit for Per Capita Income vs. Governor's Salary is
(0.2936,2.62032) Since all the values are positive we can
conclude that there is a positive relationship between Per
Capita Income vs. Governor's Salary