Business Analysis: Hypotheses Testing, ANOVA, and Chi-Square
Added on 2023-06-09
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Business Analysis
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Course Name
Institution Affiliation
Student Name
Course Name
Institution Affiliation
Business Analysis
Question One
a. Stating hypotheses to determine if there is a significant difference between the
salaries of males and female professors.
The hypotheses will be
H0 : μ0=μ1
H1=μ0 ≠ μ1
b. Calculation of test statistic to test your hypotheses
Due to an unequal sample size of the two variables, t-statistics (unpaired test), equal
variance, will be conducted.
t= ( X −Y )
√ ( sx
2
n1
+ sy
2
n2 )
where , s x
2∧s y
2 are the variance of X∧Y respectively , X∧Y are the means of X
Y are the means of X∧Y respectively , n1∧n2 are the sample ¿ X a nd Y
Note the degree of freedom in this case will be given by df =n1 +n2−2=10+11−2=19
No. Male professor salary(000s)(X) Male professor salary(000s)(Y)
1 75.4 74.8
2 78.5 79.5
3 78.7 76.5
4 81.3 74.8
5 85.3 76.4
6 77.6 72.4
7 75.5 71.5
8 82.8 74.5
9 78.5 78.5
10 80.5 80.5
11 76.5
n (n1∧n2) 10 11
Mean(X ∧Y 79.41 75.99
Variance
(s¿¿ x ¿¿ 2∧s y
2)¿ ¿ 9.75 7.77
Question One
a. Stating hypotheses to determine if there is a significant difference between the
salaries of males and female professors.
The hypotheses will be
H0 : μ0=μ1
H1=μ0 ≠ μ1
b. Calculation of test statistic to test your hypotheses
Due to an unequal sample size of the two variables, t-statistics (unpaired test), equal
variance, will be conducted.
t= ( X −Y )
√ ( sx
2
n1
+ sy
2
n2 )
where , s x
2∧s y
2 are the variance of X∧Y respectively , X∧Y are the means of X
Y are the means of X∧Y respectively , n1∧n2 are the sample ¿ X a nd Y
Note the degree of freedom in this case will be given by df =n1 +n2−2=10+11−2=19
No. Male professor salary(000s)(X) Male professor salary(000s)(Y)
1 75.4 74.8
2 78.5 79.5
3 78.7 76.5
4 81.3 74.8
5 85.3 76.4
6 77.6 72.4
7 75.5 71.5
8 82.8 74.5
9 78.5 78.5
10 80.5 80.5
11 76.5
n (n1∧n2) 10 11
Mean(X ∧Y 79.41 75.99
Variance
(s¿¿ x ¿¿ 2∧s y
2)¿ ¿ 9.75 7.77
Business Analysis
t= ( 79.41−75.99 )
√ ( 9.75
10 + 7.77
11 ) = 3.42
√ 1.6818
¿ 3.42
1.2968 =2.636
Thus tcomputed=2.636
c. Specification and justification of an appropriate probability for committing
Type I error (α)
The probability of committing type I error will be 5%, which is the probability of
incorrectly rejecting the null hypothesis. This will be attained at the significance level
of 95%, which is the probability of rejecting or accepting the null hypothesis
accurately.
To make the decision, in this case, the critical value of need to determine from
t-tables
t0.05 ( 19 df ) ( twotailed ) =1.729
d. Reporting of the decision and clearly explain your result.
The tcomputed=2.636 ¿ tα=1.729 , therefore, null hypothesis,H0 t h e t h e : μ0 =μ1 will
be rejected. This suggests that there is significant diffa erence between the salaries of
males and female professors. In this case, alternative hypothe thesis will be accepted
as it’s supported by the results of the test conducted.
In conclusion, the salary of a male professor and a female professor is not the same.
t= ( 79.41−75.99 )
√ ( 9.75
10 + 7.77
11 ) = 3.42
√ 1.6818
¿ 3.42
1.2968 =2.636
Thus tcomputed=2.636
c. Specification and justification of an appropriate probability for committing
Type I error (α)
The probability of committing type I error will be 5%, which is the probability of
incorrectly rejecting the null hypothesis. This will be attained at the significance level
of 95%, which is the probability of rejecting or accepting the null hypothesis
accurately.
To make the decision, in this case, the critical value of need to determine from
t-tables
t0.05 ( 19 df ) ( twotailed ) =1.729
d. Reporting of the decision and clearly explain your result.
The tcomputed=2.636 ¿ tα=1.729 , therefore, null hypothesis,H0 t h e t h e : μ0 =μ1 will
be rejected. This suggests that there is significant diffa erence between the salaries of
males and female professors. In this case, alternative hypothe thesis will be accepted
as it’s supported by the results of the test conducted.
In conclusion, the salary of a male professor and a female professor is not the same.
Business Analysis
Question Two
Data on blood glucose level in diabetic patients; it’s a record of the blood glucose
level before the program and after the program.
a. Stating hypotheses to determine if there is a significant difference between blood
glucose levels before and after the program
The hypotheses are
H0 : μ0=μ1
H1=μ0 ≠ μ1
b. Computation of test statistic to test the hypotheses
Here the t statistic (paired test) will be applied
t= ∑ D
√ n ∑ D2− (∑ D )2
n−1
where the D=difference between X∧Y , n=sample ¿ 10
In this case, the degree of freedom will be given by n−1=10−1=9
No. Glucose level Before Program(X) Glucose level after Program (Y) D=X-Y D^2
1 268 106 162 26244
2 225 186 39 1521
3 252 223 29 841
4 192 110 82 6724
5 307 203 104 10816
6 228 101 127 16129
7 246 211 35 1225
8 298 176 122 14884
9 231 194 37 1369
10 185 203 -18 324
Su
m 719 80077
n 10 10
Therefore,
Question Two
Data on blood glucose level in diabetic patients; it’s a record of the blood glucose
level before the program and after the program.
a. Stating hypotheses to determine if there is a significant difference between blood
glucose levels before and after the program
The hypotheses are
H0 : μ0=μ1
H1=μ0 ≠ μ1
b. Computation of test statistic to test the hypotheses
Here the t statistic (paired test) will be applied
t= ∑ D
√ n ∑ D2− (∑ D )2
n−1
where the D=difference between X∧Y , n=sample ¿ 10
In this case, the degree of freedom will be given by n−1=10−1=9
No. Glucose level Before Program(X) Glucose level after Program (Y) D=X-Y D^2
1 268 106 162 26244
2 225 186 39 1521
3 252 223 29 841
4 192 110 82 6724
5 307 203 104 10816
6 228 101 127 16129
7 246 211 35 1225
8 298 176 122 14884
9 231 194 37 1369
10 185 203 -18 324
Su
m 719 80077
n 10 10
Therefore,
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