Answer:
As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.
Step-by-step explanation:
Here n= 20
Sample mean GPA = x`= 2.84
Standard mean GPA = u= 2.55
Standard deviation = s= 0.45.
Level of Significance.= ∝ = 0.01
The hypothesis are formulated as
H0: u1=u2 i.e the GPA of night students is same as the mean GPA of day students
against the claim
Ha: u1≠u2
i.e the GPA of night students is different from the mea GPA of day students
For two tailed test the critical value is z ≥ z∝/2= ± 2.58
The test statistic
Z= x`-u/s/√n
z= 2.84-2.55/0.45/√20
z= 0.1441
As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.
First distribute: -4h - 20 + 4h
Combine like terms: -20
The answer is -20
Perimeter = 2(L + W)
2(L + 10) = 50
2L + 20 = 50
2L = 30, L = 15
Solution: the length is 15 inches
Answer:
7/ 36
Step-by-step explanation:
Roll of 2 fair six sided die :
Sample space = 6² = 36
Sum of 7:
(1,6) ; (6,1) ; (3,4) ; (4,3) ; (2,5) ; (5,2) = 6
Probability = required outcome / Total possible outcomes
Required outcome = 7
Total possible outcomes = 36
P(obtaining a sum of 7) = 7 / 36
