Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: motor skills of a sober subject.
n= 20
X[bar]= 37.1
S= 3.7
The claim is that the average score for all sober subjects is equal to 35.0, symbolically: μ= 35.0
The hypotheses are:
H₀: μ = 35.0
H₁: μ ≠ 35.0
α: 0.01
The statistic to use, assuming all conditions are met, is a one sample t- test
[tex]t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~t_{n-1}[/tex]
[tex]t_{H_0}= \frac{37.1-35.0}{\frac{3.7}{\sqrt{20} } } = 2.34[/tex]
This test is two-tailed, meaning, the rejection region is divided in two and you'll reject the null hypothesis to low values of t or to high values of t:
[tex]t_{n-1;\alpha /2}= t_{19; 0.005}=-2.861[/tex]
[tex]t_{n-1;1-\alpha /2}= t_{19; 0.995}= 2.861[/tex]
The decision rule using this approach is:
If [tex]t_{H_0}[/tex] ≤ -2.861 or if [tex]t_{H_0}[/tex] ≥ 2.861, you reject the null hypothesis.
If -2.861 < [tex]t_{H_0}[/tex] < 2.861, you do not reject the null hypothesis.
The value is within the non rejection region, the decision is to not reject the null hypothesis.
I hope this helps!
