Question

 

17) In an experiment the value for beta has been shown to be .7503, what is the power of this experiment? 

18) A researcher did a one-tailed hypothesis test using an alpha level of .01, H0 was rejected. A colleague analyzed the same data but used a two-tailed test with α=.05, H0 was failed to reject. Can both analyses be correct? Explain your answer.

21) A psychologist would like to examine the effect of fatigue on mental alertness. An attention test is prepared which requires subjects to sit in front of a blank TV screen and press a response button each time a dot appears on the screen. A total of 110 dots are presented during a 90-minute period, and the psychologist records the number of errors for each subject. Two groups of subjects are selected. The first group (n=5) is tested after they have been kept awake for 24 hours. The second group (n=10) is tested in the morning after a full night’s sleep. The data for these two samples are as follows:

____________________________________

Awake 24 hours     Rested

____________________________________

x̅=35     x̅=24

ss=120 ss=360

____________________________________

On the basis of these data, can psychologist conclude that fatigue significantly increases errors on an attention task? Test with α=.05 and report effect size. 

Answer 1

17)  
beta =    0.7503
Power = 1- beta =   0.2497
  
18)  
Case 1 - one-tailed hypothesis with a =.01. H0 was rejected.
-> p-value <= 0.01
Case 2 - two-tailed test with α=.05, H0 was failed to reject.
p-value > 0.05

Explanation 1) (assume case1 is true)
p-value _ two-tailed test = 2 * p-value _ one tailed test
largest p-value for one-tailed test should be 0.01
Therefore, largest p-value for two-tailed test should be, p-value = 2 * 0.01 = 0.02 which is less than alpha level of 0.05
And since, p-value = 0.02 less than alpha level of 0.05, we should reject H0.
Therefore, case 2) result would not be correct.

Explanation 2) - assume case 2) is correct.
p-value > 0.05 (given)
Suppose the largest p-value for two-tailed test is 0.05,
then p-value for one-tailed should be, p-value = 0.05/2 = 0.025
which is greater than alpha level of 0.01 and we should fail to reject H0.
That means case 1) would no be correct.

19)

The Null hypothesis, there is a significant difference in the average between Awake 24 hours and rested. u1=u2.

An alternative hypothesis, the average error for a week 24 hours English in comparison to Rested. u1 < u2.

	sample 1	sample 2
n=	5.000	10.000
mean=	35.0000000	18.5200000
s= sqrT(SS/(n-1)	5.4772000	6.3246000
s^2/n	5.9999	4.0001

Sp^2
{(n1-1)*s1^2 + (n2-1)*s2^2} / {n1+n2-2}
=((5-1)*5.4772^2+(10-1)*6.3246^2)/(5+10-2)
36.92338

Test statistic, t=
(Xbar1 - Xbar2)/sqrt(Sp^2*(1/n1+1/n2))
(35-18.52)/SQRT(36.92338*(1/5+1/10))
4.9516

df= n1+n2-2
5+10-2
13.000

critical value: -t(a, df)
=-t(0.05,13)
T.INV(0.05,13)
-1.770933396

p-value
(1-P(T 1-P(T<4.9516)
T.DIST.RT(4.9516,13)
0.000132343

With t=4.95, p<5%, i reject ho and conclude that the average error for a week 24 hours English in comparison to Rested. u1 < u2.