Answer:
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
Step-by-step explanation:
We formulate null and alternate hypotheses are
H0 : u1 < u2 against Ha: u1 ≥ u 2
Where u1 is the group tested after they were awake for 24 hours.
The Significance level alpha is chosen to be ∝ = 0.05
The critical region t ≥ t (0.05, 13) = 1.77
Degrees of freedom is calculated df = υ= n1+n2- 2= 5+10-2= 13
Here the difference between the sample means is x`1- x`2= 35-24= 11
The pooled estimate for the common variance σ² is
Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]
= 1/13 [ 120²+360²]
Sp = 105.25
The test statistic is
t = (x`1- x` ) /. Sp √1/n1 + 1/n2
t= 11/ 105.25 √1/5+ 1/10
t= 11/57.65
t= 0.1908
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
40/11 = 3.63
hope this helps
Answer:
The answer is 325.3
Step-by-step explanation:
I hope this helped I used a calculator sorry if I'm late and I hope this isn't wrong
53. The ratio of apple over orange = 3:2 The total number of fruits are 60 pieces. Now, find the total number of apples and total number of oranges. => first, let’s add 2 + 3 = 5 Thus, we have 5 division for 60 pieces fruits, => 60 / 5 => 12 let’s solve for the apple => 12 x 3 = 36 Now. The orange => 12 * 2 = 24 => 36 + 24 = 60 Thus the ratio is 36 : 24
