A team of sports scientists want to know if having a dietitian as part of the...

A team of sports scientists want to know if having a dietitian as part of the staff is better at producing divisional basketball champions than current ad-libitum (as desired) nutritional practices. Of the ad-libitum, 12 were champions and 18 were non-champions. Of the dietician-tailored, 7 were champions and 8 were non-champtions. What can they conclude?

Choose the most correct answer.

Select one:

a. After performing a 2 x 2 chi square test, no significant difference between champion and non-champion basketball teams was found between the two dietary approaches (χ2 = 0.18, df = 1, p < .05).

b. After performing a 2 x 2 chi square test, no significant difference between champion and non-champion basketball teams was found between the two dietary approaches (χ2 = 0.18, df = 1, p > .05).

c. After performing a 2 x 2 chi square test, no significant difference between champion and non-champion basketball teams was found between the two dietary approaches (χ2 = 0.01, df = 1, p < .05).

d. After performing a 2 x 2 chi square test, a significant difference between champion and non-champion basketball teams was found between the two dietary approaches (χ2 = 0.01, df = 1, p < .05).

e. After performing a 2 x 2 chi square test, a significant difference between champion and non-champion basketball teams was found between the two dietary approaches (χ2 = 3.84, df = 1, p < .05).

Solutions

Expert Solution

create and contingenct table with 2 rows and 2 columns

with chisq.test fucntion perform chi sq test on contingency table

Ho:no asscociation between champion and non-champion basketball teams was found between the two dietary approaches

Ha:there is an association between champion and non-champion basketball teams was found between the two dietary approaches

Rcode:

R1 = c(12,18)
R2 = c(7,8)
rows = 2
contingencytable = matrix(c(R1, R2),
nrow=rows,
byrow=TRUE)
colnames(contingencytable) = c("champions ", "non-champions") # Naming the rows and columns
rownames(contingencytable) = c("Diet1","Diet2") # columns is optional.

contingencytable

chisq.test(contingencytable,correct=FALSE)

Output:

Pearson's Chi-squared test

data: contingencytable
X-squared = 0.18219, df = 1, p-value = 0.6695

X-squared = 0.18219, p=0.6695,p>0.05

Fail to reject Ho

b. After performing a 2 x 2 chi square test, no significant difference between champion and non-champion basketball teams was found between the two dietary approaches (χ2 = 0.18, df = 1, p > .05)

orchestra answered 2 years ago

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