Question

For each of the following sets of results, compute the appropriate test statistic, test the indicated alternative hypothesis, and compute the effects size(s) indicating their magnitude:

set	hypothesis	1	2	D	n	α
a)	μ1 ≠ μ2	32.6	32.1	4.2	24	0.05
b)	μ1 > μ2	101.9	95	10	27	0.01
c)	μ1 < μ2	74.6	70.2	6.6	24	0.10

a)
Compute the appropriate test statistic(s) to make a decision about H₀.
critical value? =  ; test statistic? =
Decision?:  ---Select--- Reject H0 Fail to reject H0

Compute the corresponding effect size(s) and indicate magnitude(s).
d? = ; Magnitude?:  ---Select--- na trivial effect small effect medium effect large effect
r^2? = ; Magnitude?:  ---Select--- na trivial effect small effect medium effect large effect

b)
Compute the appropriate test statistic(s) to make a decision about H₀.
critical value? =  ; test statistic? =
Decision?:  ---Select--- Reject H0 Fail to reject H0

Compute the corresponding effect size(s) and indicate magnitude(s).
d? = ; Magnitude?:  ---Select--- na trivial effect small effect medium effect large effect
r^2? = ; Magnitude?:  ---Select--- na trivial effect small effect medium effect large effect

c)
Compute the appropriate test statistic(s) to make a decision about H₀.
critical value? =  ; test statistic? =
Decision?:  ---Select--- Reject H0 Fail to reject H0

Compute the corresponding effect size(s) and indicate magnitude(s).
d? = ; Magnitude?:  ---Select--- na trivial effect small effect medium effect large effect
r^2? = ; Magnitude?:  ---Select--- na trivial effect small effect medium effect large effect

Answer 1

Answer:

Given that,

For each of the following sets of results, compute the appropriate test statistic, test the indicated alternative hypothesis, and compute the size of the effect (s) indicating their magnitude:

(a).

(Two-tailed test)

The population standard deviation()=4.2

The sample size (n)=24

The sample mean ()=32.6

The critical value is = 1.96

The standard error (SE)=

=4.2 /4.899

SE=0.8573

The Z- test statistic is,

=0.5/0.8573

Z=0.5832 (Approximately)

The p-value at Z=0.5832 and is,

The P-Value is 0.5598.

The result is not significant at p < 0.05.

Decision: Do not reject H0.

=0.5/ 4.2

d=0.1191 (Approximately)

=0.0142/(0.0142+4)

=0.0142/4.0142

r^2=0.0035

(b).

(Two-tailed test)

The population standard deviation()=10

The sample size (n)=27

The sample means ()=101.9

The critical value is = 2.58

The standard error (SE)=

=10 /5.1962

SE=1.9245

The Z- test statistic is,

=6.9 /1.9245

Z=3.5854 (Approximately)

The p-value at Z=3.5854 and is,

The P-Value is 0.00034.

The result is significant at p < 0.01.

Decision: Reject H0.

=6.9/ 10

d=0.69 (Approximately)

=0.4761/(0.4761+4)

=0.4761/4.4761

r^2=0.1064

(c).

(Two-tailed test)

The population standard deviation()=6.6

The sample size (n)=24

The sample means ()=74.6

The critical value is = 1.65

The standard error (SE)=

=6.6 /4.899

SE=1.3472

The Z- test statistic is,

=4.4/1.3472

Z=3.2660 (Approximately)

The p-value at Z=3.2660 and is,

The P-Value is 0.0011.

The result is significant at p < 0.10.

Decision: Reject H0.

=4.4 / 6.6

d=0.67 (Approximately)

=0.4489/(0.4489+4)

=0.4489/4.4489

r^2=0.1009

**Please comment on any doubt.