Peanut allergy is increasingly common in Western countries. A study enrolled infants with a diagnosed peanut sensitivity and randomly assigned them to either avoid or regularly consume peanuts until 60 months of age. At the end of the study, 18 of the 51 children who had avoided peanuts were still allergic to peanuts. In contrast, 5 of the 47 children who had regularly consumed peanuts were still allergic to peanuts. We want to estimate how much difference the two approaches make by using a confidence interval for the difference between the proportions with a peanut allergy at 60 months in the two populations. (a) Which of the choices explains why we cannot use the large sample method?
i. There are exactly 10 successes and failures in each sample. Therefore, we should not use the large-sample confidence interval.
ii. There are more than 10 successes and failures in each sample. Therefore, we should not use the large-sample confidence interval.
iii. There are 10 successes and failures in each sample. Therefore, we should be able to use the large-sample confidence interval.
iv. There are not 10 successes and failures in each sample. Therefore, we should not use the large-sample confidence interval.
(b) Obtain the lower value and upper value for the appropriate 95% plus four confidence interval.

B. Does early exposure to peanuts in small amounts or complete avoidance of peanuts better reduce peanut allergy in children with a known allergy to peanuts?

Step-by-step explanation:

Remember we are told there's already a question whether pediatricians should recommend early exposure or avoidance. Thus, the research question should answer these two issues (early exposure or avoidance).

Hence, the question we are asking is: Does early exposure to peanuts in small amounts or complete avoidance of peanuts better reduce peanut allergy in children with a known allergy to peanuts?

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.  

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it