Question

A psychologist is interested in the effect that no sleep has on individuals’ abilities to perform motor tasks (physical abilities). She recruits a sample of n = 25 college students to serve as the control group, which is allowed to sleep up to 10 hours. A second sample of n = 25 college students is selected and stays up all night. The following morning, the students attempt to make 20 free throw shots with a basketball, the number of shots made are recorded. The data for the two groups is listed below:

Sleep
n₁ = 25
M₁ = 12
SS₁ = 916

No sleep
n₂ = 25
M₂ = 8
SS₂ = 620

Determine whether lack of sleep had a significant effect on motor tasks using α =.05 test of significance by answering the following questions:

1.Write the null hypotheses in words.

2.Write the alternative hypotheses in words.

3. What is the critical t value(s) that sets off the critical boundary or boundaries?

4.What is the pooled variance?

5. What is the estimated standard error (standard error of the mean difference)?

6.What is the calculated t statistic for our sample?

7.What is your decision (reject or fail to reject the null hypothesis)?

8.Explain in words the result of this t-test.

9. Compute Cohen’s d (even if not significant) to measure the size of the treatment effect.

10. Interpret Cohen’s d.

Answer 1

Given that,
mean(x)=12
standard deviation , s.d1=6.177
number(n1)=25
y(mean)=8
standard deviation, s.d2 =5.082
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.01
since our test is two-tailed
reject Ho, if to < -2.01 OR if to > 2.01
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (24*38.1553 + 24*25.8267) / (50- 2 )
s^2 = 31.991
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=12-8/sqrt((31.991( 1 /25+ 1/25 ))
to=4/1.5998
to=2.5004
| to | =2.5004
critical value
the value of |t α| with (n1+n2-2) i.e 48 d.f is 2.01
we got |to| = 2.5004 & | t α | = 2.01
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 2.5004 ) = 0.0158
hence value of p0.05 > 0.0158,here we reject Ho
ANSWERS
---------------
1.
null, Ho: u1 = u2
2.
alternate, H1: u1 != u2
3.
critical value: -2.01 , 2.01
4.
pooled variance = 31.991
5.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((38.155/25)+(25.827/25))
= 1.6
6.
test statistic: 2.5004
7.
decision: reject Ho
p-value: 0.0158
8.
we have enough evidence to support the claim that whether lack of sleep had a significant effect on motor tasks
9.
cohen's d size = mean difference / pooled standard deviation
pooled standard deviation = sqrt(((n1-1)*s.d1^2 + (n2-1)*s.d2^2)/(n1+n2-2))
pooled standard deviation = sqrt(((24)*6.177^2 +(24)* 5.082^2)/(50-2)) =5.656
cohen's d size = (12-8)/5.656 =0.707
10.
medium Effect is 0.50 to 0.80
small effect is 0.20 to 0.50
large effect is greater than 0.80