An experiment was conducted to explore the effects of extended periods of sensor

Question

An experiment was conducted to explore the effects of extended periods of sensory de- privation on alpha wave frequencies in EEG (electro encephalogram) representations of brain activity. Higher alpha wave frequencies are generally associated with lower levels of stress and anxiety Twenty volunteers, all of whom were inmates in a Canadian prison, were randomly divided into two groups of ten inmates. The members of one group were placed in solitary confine ment and were allowed no contact with other inmates. The members of the other group were allowed to remain in their usual cells and were allowed normal contact with other inmates. After seven days under these conditions, alpha wave frequencies were measured via an EEC, for each inmate. Note: Before the experiment began the measured alpha wave frequencies of all 20 inmates were reasonably similar. The data values (alpha-wave frequencies, in Hertz (Hz)) are given below. (source: P. Gendreau et al. (1972), Journal of Abnormal Psychology, 79, 54-59, as described in De Veaux and Velleman (2004)) Part 1: Some basics. 1. Clearly define two relevant population means, 1 and 2, with the subscript 1 indicating "confined" and the subscript 2 indicating "not confined" 2. State what you believe to be an appropriate research hypothesis relating to 2 Provide a clear and complete statement of this hypothesis in words in the context of this problem and explain why your hypothesis is appropriate The data. alpha-wave frequencies solitary confinement group: 9.0 9.7 9.2 9.9 9.3 10.3 10. 10.9 9.5 9.6 normal conditions group: 9.6 10.70.710.911 10.3 10. 010.5 11.2

Explanation / Answer

M=Mean1=9.78

M=mean2=10.58

then for treatment1

diff(X-M)

-0.78   
-0.58
-0.48
-0.28
-0.18
-0.08
0.12
0.52
0.62
1.12

then square the given column and sum all the value we get SS=3.22

For Treatment two:

M=mean2=10.58

diff(X-M)

-0.98
-0.28
-0.18
-0.18
-0.08
0.12
0.12
0.32
0.52
0.62

then square the given column and sum all the value we get SS=1.90

Difference Scores Calculation

H0 :Mean1=Mean2 (both the group have equal level of alpha wave frequency)

H1:mean1<mean2  ( group1 have less alpha wave frequency than group two)

alpha=0.01

Treatment 1

N1: 10
df1 = N - 1 = 10 - 1 = 9
M1: 9.78
SS1: 3.22
s21 = SS1/(N - 1) = 3.22/(10-1) = 0.36


Treatment 2

N2: 10
df2 = N - 1 = 10 - 1 = 9
M2: 10.58
SS2: 1.9
s22 = SS2/(N - 1) = 1.9/(10-1) = 0.21


T-value Calculation
pooled variance :
s2p = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22) = ((9/18) * 0.36) + ((9/18) * 0.21) = 0.28

s2M1 = s2p/N1 = 0.28/10 = 0.03
s2M2 = s2p/N2 = 0.28/10 = 0.03

t = (M1 - M2)/(s2M1 + s2M2) = -0.8/0.06 = -3.36

t=-3.36

p value=0.001756

tha result is significant as <p value.

hence we reject null hypothesis.

and can say that gruop 1 have less alpha wave frequency then group two.

that implice that the member which kept in normal conditon have high alpha wave frequency hence have low stress and anxity level.

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