Question

Dr. Pepper wants to examine if a training program has an effect on weekly exercise. College students exercise a mean of 2.3 days a week with a variance of 0.81 days. Dr. Pepper's sample of 26 students exercise a mean of 2.1 days a week. What can be concluded with an α of 0.05?

a) What is the appropriate test statistic?
---Select--- na, z-test, one-sample t-test, independent-samples t-test, related-samples t-test

b)
Population:
---Select--- individuals exposed to the training program, (exercise), students, days in the week, or college students
Sample:
---Select--- individuals exposed to the training program, (exercise), students, days in the week, or college students

c) Obtain/compute the appropriate values to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value = ________; test statistic = _________
Decision:  ---Select--- Reject H0, Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[ _______ ,___________ ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and select "na" below.
d = ___________ ;   ---Select--- na, trivial effect ,small effect , medium effect, large effect
r² = _____________;   ----Select--- na, trivial effect ,small effect , medium effect, large effect

f) Make an interpretation based on the results.

A) Individuals that went through the training program did significantly more exercise than college students.

B) Individuals that went through the training program did significantly less exercise than college students.    

C) The training program has no significant effect on weekly exercise.

Answer 1

a) Appropriate test statistic : One sample t test

As we are comparing a single sample mean with the population mean , and also the sample size is small , thus we use one sample t test .

b) Population : College students

Sample : Individuals exposed to training program

c) The null and alternative hypothesis

The null and alternative hypothesis

2.3

Test statistic

Given , 2.1

Since sample variance(s²) is not given , we use population variance ()

s²= 0.81 , s =0.9

  

= 0.11  

df = n-1 = 26- 1=25

The two  tailed critical value of t at 0.05 , with df =25 is tc = 2.060

Critical value= 2.060

Test statistic =0.11

Since test statistic < Critical value .

Fail to Reject H0

d)

95% confidence interval for population mean

For 95% confidence with df =25 , tc = 2.060

95% confidence interval for population mean

= (1.74 , 2.46)

e) Cohen's d is given by

  

Effect is small

Note : ignore the negative sign

Note : Cohen's d

small effect :0.2

medium effect : 0.5

large effect : 0.8

Effect size

r^2 = na

Note : for one sample t test we use cohen's d

f) C) Training program has no significant effect on weekly exercise

Note : As test statistic < Critical value ., the difference is not significant . There is not sufficient evidence to conclude that training program has significant effect on weekly exercise