-sqrt(36) = -6
Hope this helps :)
The 'x' coordinate of the midpoint is the average of the 'x'
Answer:
H0 : μ1 - μ2 = 0
H1 : μ1 - μ2 ≠ 0
-1. 34
0.1837
Step-by-step explanation:
Full time :
n1 = 125
x1 = 2.7386
s1 = 0.65342
Part time :
n2 = 88
x2 = 2.8439
s2 = 0.49241
H0 : μ1 - μ2 = 0
H1 : μ1 - μ2 ≠ 0
Test statistic :
The test statistic :
(x1 - x2) / sqrt[(s1²/n1 + s2²/n2)]
(2.7386 - 2.8439) / sqrt[(0.65342²/125 + 0.49241²/88)]
−0.1053 / sqrt(0.0034156615712 + 0.0027553)
-0.1053 /0.0785554
= - 1.34
Test statistic = - 1.34
The Pvalue :
Using df = smaller n - 1 = 88 - 1 = 87
Pvalue from test statistic score ;
Pvalue = 0.1837
Pvalue > α ; We fail to reject the null and conclude that the GPA does not differ.
At α = 0.01 ; the result is insignificant
Answer:
I can't understand so don't worry
