Question 2 Background: PKC and Anxiety in Mice Researchers investigated the effe

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Question 2 Background: PKC and Anxiety in Mice Researchers investigated the effect that the enzyme protein kinase C (PKC) has on anxiety in mice. The genotype for a particular gene in a mouse (or a human) consists of two alleles (copies) of each chromosome, one each from the father and mother. The investigators in the study separated the mice into three groups. In Group a, neither of the mice's alleles for PKC produced the enzyme. In Group b, one of the two alleles for PKC produced the enzyme and the other did not. In Group c, both PKC alleles produced the enzyme. To measure the anxiety in the mice, scientists measured the time (in seconds) the mice spent in the "open-ended" sections of an elevated maze. It was surmised that mice spending more time in open-ended sections exhibit decreased anxiety. The researchers will to conduct a test, using a 5% level of significance, to assess whether the population mean time spent (in seconds in the open ended sections of the maze was the same for all three groups. Question 2 Subquestions 2.a Which of the following graphical displays should be used to check the ANOVA assumption that each population has a normal distribution? point(s) One QQ plot (for all the data combined) Three QQ plots (one for each group) One box plot (for all the data combined) Three box plots (one for each group) 2.b Levene's test was conducted using R. Here is the output produced: point(s) Levene's Test for Homogeneity of Variance (center mean) Df F value Pr (>F) group 2 0.58 0.56 Which of the following is the appropriate null hypothesis that the Leven's test is designed to assess?

Explanation / Answer

2a

b) Three QQ plots (one for each group)
It checks the normalty of each group

2b
answer c
Levene's test check for the equality of population

variances

2c

answer a
since the p-value > 0.05, we cannot reject h0.
therefore the variances are equal as per H0.

2d

answer a
PR( >F) = Probability of observing f-stat greater

than 4.923 is 0.12

2e

answer b
since p-value < 0.05, reject h0. equality of means

is rejected.
Meaning - atleast one of the populaton means is

different from the other groups.

all the other answers are bonus answers, as i can

only answer as per the terms of chegg

2f

mean square is the sample variance which is an

estimate of pop variance
answer c - 335.2

2g
CI = mean +/- z0.95* sd/sqrt(n)
= mean +/-1.96*sd/ sqrt(n)

2h
answer c - group b versus c

because when the lower and higher bounds include

zero or when they have opposite signs, then there

is significant difference

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