Keenan et al 2001 used anesthesia to investigate which brain

Keenan et al. (2001) used anesthesia to investigate which brain hemisphere is involved in self-recognition. Ten subjects were randomly assigned to two groups. The left hemisphere was anesthized in one group, whereas the right hemisphere was anesthized in the other group. Each subject was then shown a picture generated by averaging (\"morphing\") images of the face of a famous celebrity (e.g., Marilyn Monroe) and their own face, and told to remember the picture. After recovery from anesthesia, patients were presented with two pictures and asked to choose the one they had been shown earlier while under anesthesia. The two pictures were the original two images from which the morphed image had been generated (i.e., \"self\" and \"celebrity,\" but seperate this time). All five patients whose left hemisphere had been inactivated chose the picture of self. Four of the five patients whose right hemisphere had been anesthetized chose the celebrity picture, instead (the fifth chose self). State what test you would use to determine whether the treatment (left vs. right hemisphere anesthesia) influenced recognition of self versus celebrity. Explain why you would use this test. (Carry out the test if possible).

Solution

Here paired test can be used.

Because in a group the group is divided into two groups. And each group given a treatment different from other. Hence this can be called paired test .

T test is better as sample size is very small.

After conducting the test the results are recorded.

If p1 is the proportion of identifying self, then p1 = 1

while p2 for second group it is 0.2

Sample size =5 each.

McNemar Chi Square test statistic ?² 1

McNemar test statistic with Yates\'s correction ?² 0.25

Difference between test -0.1

Odds ratio % Infty

95% Confidence Interval upto infinity

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As degree of freedom =1 and chi square =1

p =0.3173

As p >0.05, accept null hypothesis.

By conventional criteria, this difference is considered to be not statistically significant

NOte: This may be due to small sample. Hence try to get sample size more to get more accuracy.