7.7 A researcher performs an experiment to determine if calcium supplements de-

Question

7.7 A researcher performs an experiment to determine if calcium supplements de- crease the probability of osteoporosis in lab mice. Seventy-four mice are assigned to the calcium treatment group and 226 are assigned to a control group. Yes No Calcium 74 218 291 No Calcium 8 (a) The researcher tells you that a chi-square approximation cannot be used to approximate the two-tailed Fisher's exact test p-value because one of the cells of the above table is 1. Is this reasoning correct? Why or why not? (b) Compute the p-value from Fisher's exact test to test the null hypothesis that the probability of osteoporosis is the same for mice in the two groups versus the alternative hypothesis that the probability of osteoporosis is lower in the calcium group. (c) You later learn that each mouse was also identified by gender. Of the 300 mice in the study, 225 were male (25 in the treatment group and 200 in the control group). Only 2 of the male mice suffered osteoporosis, and both of these mice were in the control group. Using this additional information, test the null hypothesis that the probability of osteoporosis for female mice is the same in

Explanation / Answer

SolutionA:

Expected freq =row total*column total/grand total

For (1,1)--1 row 1 st column which is 1 Expecetd freq is

=74*9/300=2.22

for(1,2)

E(1,2)=74*291/300=71.78

E(2,1)=

226*9/300=6.78

E(2,2)

=226*291/300=219.22

to use chi sq test

Expected Frequencies of all cells>=5

Here for firts cell

Expected Freq=2.22<5

We cannot use Chi sq test

Solutionb:

Fischer's exact test is exact because it guarantees the alpha rate,regardless of the sample size

Ho:probability of :osteoporosis is the same for the two groups

H1:probability of osteoporosis is lower in the calicim group

level of significance=0.05

Rcode is :

mat <- matrix(c(1,8,73,218), nrow=2,dimnames=list(c("Calcium","No-calcium"),c("Osteoporosis-Yes","Osteoporosis-No")))
mat
fisher.test(mat,alternative="less")

Output:

data: mat
p-value = 0.3044
alternative hypothesis: true odds ratio is less than 1
95 percent confidence interval:
0.000000 2.305856
sample estimates:
odds ratio
0.3742004

p value=0.3044

p>0.05

Fail to reject Null hyppothesis

Conclusion:

there is no sufficient statistical evidence at 5% level of significance to support the claim that the probbaility of osteoporosis is lower in the calcium group.

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