Question

A health psychologist knew that corporate executives in general have an average score of 2.55 with a standard deviation of 0.5 on a stress inventory and that the scores are normally distributed. In order to learn whether corporate executives who exercise regularly have different stress scores, the psychologist measured the stress of 30 exercising executives and found them to have a mean score of 2.76. a. Using the five steps of hypothesis testing, was the group of 30 exercising executives different from executives in general? (Use the .05 significance level.) Be sure to sketch the distributions involved. Be sure to fully state the five steps and show ALL calculations/work. b. Additionally, explain the logic of what you did to a person who understands hypothesis testing for studies in which the sample consists of a single individual but is unfamiliar with hypothesis testing involving a sample of more than one individual. (Be sure your explanation includes a discussion of the distribution of means--what it is, what it is made of, the concept of how it is created, how itʹs characteristics are created, its role in the overall hypothesis testing process.)

Answer 1

Kindly check explanation

Explanation:

Given the following :

Population mean (μ) = 2.55

Population standard deviation (σ) = 0.5

Sample size (n) = 30

Sample mean (x) = 2.76

α = 0.05

STEP 1:

Stress score in general executive (s1)

Stress score in exercising executive (s2)

Null : s1 = s2

Alternative : s1 < s2

STEP 2:

Shape of distribution = normal

Population mean (μ) = 2.55

Population standard deviation (σ) = 0.5

Sample size (n) = 30

Sample mean (x) = 2.76

α = 0.05

Decision rule :

α = 0.05 which corresponds to a t score (t0) ;

df = n - 1 = 30 - 1= 30 at 0.05 = 1.699

If :

(Test statistic (t) > t0) ; reject the Null

(right tailed test)

Test statistic (t) :

(x - μ) / (σ/√n)

(2.76 - 2.55) / (0.5/√30)

0.21 / 0.0913

= 2.30

t > t0

2.30 > 1.699

t is more extreme than t0

Hence, reject the null at α = 0.05