Question

Clinicians hypothesize there is a difference in learning self-care skills between two techniques. For 4 months the clinicians worked with mentally impaired individuals in which half were taught self-care skills through video imitation and the other half through physical guidance. At the end of the 4 months, individuals were asked to independently perform the self-care task and were rated on the level of required assistance where a higher score indicates more assistance. What can be concluded with an α of 0.05? The rating data are below.

video imitation	physical guidance
10 10 17 11 11 14 11 14 16	15 13 16 11 19 21 16 14 20

a) What is the appropriate test statistic?
---Select one--- (na, z-test, One-Sample t-test, Independent-Samples t-test, Related-Samples t-test)
b)
Condition 1:
---Select one--- (physical guidance, 4 months, mentally impaired, video imitation, self-care skills)
Condition 2:
---Select one--- (physical guidance, 4 months, mentally impaired, video imitation, self-care skills)
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 one--- (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/or select "na" below.
d =  ;  ---Select one--- (na, trivial effect, small effect, medium effect, large effect)
r² =  ;  ---Select one--- (na, trivial effect, small effect, medium effect, large effect)
f) Make an interpretation based on the results.

a. Self-care skills improve when learned through physical guidance.

b. Self-care skills improve when learned through video imitation.    

c. There is no significant self-care difference between video imitation and physical guidance.

Answer 1

Sr. No	video imitation	physical guidance
1	10	15
2	10	13
3	17	16
4	11	11
5	11	19
6	14	21
7	11	16
8	14	14
9	16	20
Total	114.000	145.000
Mean	12.667	16.111
Var	7.000	11.111
n	9.000	9.000

Mean =

Var

a) What is the appropriate test statistic?
Independent-Samples t-test

This is because the test is to compare two population means where the samples selected ae not same for both the data sets. Hence they are indpendent.

b)
Condition 1:
video imitation

Condition 2:
physical guidance

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.)

The cliician say that there is a difference not indicating a positive or a negative one. Hence, it is a two tailed test where the null hypothesis difference between means =0

(There is no difference between the ratings)

(There is a difference between the ratings)

We are testing at 0.05
critical value =

=

Critical value = 2.12 .........using t-dit tables with p = 2.5% and df = 9 + 9 - 2 = 16

; test statistic =

Where the null difference =

Where the pooled variance =

= 9.0556

Test stat =

Test Stat = -2.428

Since |Test stat| > critical value
Decision:  Reject H0

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

95% confidence interval for difference between the population mean is given by

We have all the values we need

(-6.4517,-0.4372)

e) Compute the corresponding effect size(s) and indicate magnitude(s).

Effect size =

=

Effect size = 1.1446 we take the absolute value
since effect size > 0.8

Large effect

f) Make an interpretation based on the results.

Well since the ratings were more for physical guidence we can say that

a. Self-care skills improve when learned through physical guidance.

Although for checking this we need to conduct a 1-tailed test.

c. There is no significant self-care difference between video imitation and physical guidance.

No significance is when we do not rejec the hypothesis.