Question

A test of sobriety involves measuring the subject's motor skills. Twenty randomly selected sober subjects take the test and produce a mean score of A with a standard deviation of B. At the 0.01 level of significance, test the claim that the true mean score for all sober subjects is equal to 35.0. Use the critical value method of testing hypotheses.

Answer 1

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

$t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~t_{n-1}$

$t_{H_0}= \frac{37.1-35.0}{\frac{3.7}{\sqrt{20} } } = 2.34$

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:

$t_{n-1;\alpha /2}= t_{19; 0.005}=-2.861$

$t_{n-1;1-\alpha /2}= t_{19; 0.995}= 2.861$

The decision rule using this approach is:

If $t_{H_0}$ ≤ -2.861 or if $t_{H_0}$ ≥ 2.861, you reject the null hypothesis.

If -2.861 < $t_{H_0}$ < 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!