Seven students volunteered for a comparison of study guides for an advanced course in mathematics. They were randomly assigned, four to study guide A and three to study guide B. All were instructed to study independently. Following a two-day study period, all students were given an examination about the material covered by the guides, with the following results:Study Guide A scores: 68; 77; 82; 85Study Guide B scores: 53; 64; 71Perform a randomization test by listing all possible ways that these students could have been randomized to two groups. There are 35 ways. For each outcome, calculate the difference between sample averages. Finally, calculate the two-sided p-value for the observed outcome

Step-by-step explanation:

From the given question; we can use the R software to program the combination function that generates all the combinations.

options(digits =2(

scores<- c(68,77,82,85,53,64,71)

groupA <- combn(scores,4)

groupB <- apply(groupA,2, function(x) scores[!   (scores  %in%  x)  ]  )

colnames(groupA) <- colnames(groupB) <- paste("G", 1:35, sep"")

The accompanying 35 groupings (G1 to G35) contain all potential ways these understudies can be randomized under the null hypothesis

Group A

Group B

The accompanying data below computes the distinctions for each group:

The two-sided p-value is the extent of contrasts between test midpoints as large or bigger in supreme value than the primary group. The cat function makes the outcomes simpler to peruse.

p <- sum (aba(difference)>=difference[1])/35

cat(p)

= 0.086

a

b

d

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;