The assessment of diet is an important exposure for many disease outcomes. However, there is often much imprecision in dietary recall. In one study, 70-to 79-year-old women were asked about the preschool diet of their children (ages 2−4) using a food frequency questionnaire (FFQ). A unique aspect of the study is that simultaneous diet record data exist on the same children recorded in real time by their mothers when they were ages 2−4 and their mothers were 20 to 40 years old. The data in Table 11.36 were available on average servings of margarine per week. IDFFQDR34070399 70.546600502005410055472.55587360570.5611213.761802.5653214.170778.5The Pearson correlation between intake from the two re-cording methods was 0.448. Assume that FFQ and DR margarine intake are normally distributed.Question: Provide a 95% confidence interval for ρ. #95% CI for zr <- 0.448#Fisher's transformation of rz <- 0.5*log((1+r)/(1-r)); zz + c(-1,1)*qnorm(0.975)*sqrt(1/(n-3))95% CI for z = (-0.1711261, 1.1355166)#95% CI for ρ rlower <- (exp(2*-0.1711261)-1)/(exp(2*-0.1711261)+1); rlowerrupper <- (exp(2*1.1355166)-1)/(exp(2*1.1355166)+1); rupperMy answer: 95% CI = (-0.169475, 0.8128989)Is my answer correct?

The dataset is given as: ID FFQ DR 340 7 0 399 7 0.5 466 0 0 502 0 0 541 0 0 554 7…

Answered: The assessment of diet is an important…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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The assessment of diet is an important exposure for many disease outcomes. However, there is often much imprecision in dietary recall. In one study, 70-to 79-year-old women were asked about the preschool diet of their children (ages 2−4) using a food frequency questionnaire (FFQ). A unique aspect of the study is that simultaneous diet record data exist on the same children recorded in real time by their mothers when they were ages 2−4 and their mothers were 20 to 40 years old. The data in Table 11.36 were available on average servings of margarine per week.

ID	FFQ	DR
340	7	0
399	7	0.5
466	0	0
502	0	0
541	0	0
554	7	2.5
558	7	3
605	7	0.5
611	21	3.7
618	0	2.5
653	21	4.1
707	7	8.5

The Pearson correlation between intake from the two re-cording methods was 0.448. Assume that FFQ and DR margarine intake are normally distributed.

Question: Provide a 95% confidence interval for ρ.

#95% CI for z

r <- 0.448

#Fisher's transformation of r
z <- 0.5*log((1+r)/(1-r)); z
z + c(-1,1)*qnorm(0.975)*sqrt(1/(n-3))

95% CI for z = (-0.1711261, 1.1355166)
#95% CI for ρ
rlower <- (exp(2*-0.1711261)-1)/(exp(2*-0.1711261)+1); rlower
rupper <- (exp(2*1.1355166)-1)/(exp(2*1.1355166)+1); rupper

My answer: 95% CI = (-0.169475, 0.8128989)

Is my answer correct?

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

The dataset is given as:

$I D$	$F F Q$	$D R$
340	7	0
399	7	0.5
466	0	0
502	0	0
541	0	0
554	7	2.5
558	7	3
605	7	0.5
611	21	3.7
618	0	2.5
653	21	4.1
707	7	8.5

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I thought CI around Pearson's r is not symmetrical and we would need to transform it to z, calculate the CI then undo Fisher's transformation to obtain the population correlation coefficient CI?

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