E7_Describe_answers_v3

The cell size is 22. iii. Now refer to page 158, paragraph 2 in the right column of Kohn et al. The authors report that “Among low income, food-insecure youth, food assistance participation was not associated with … high waist circumference or categorical weight status for any specification of food assistance in the fully adjusted models… (data not shown).” Based on your answers to part d.i. and d.ii. above and guidance offered in the E.7 video, what is a plausible explanation for the lack of association between food assistance participation and high waist circumference or the categorical body size measures among food insecure boys? A plausible explanation for the lack of association between food assistance participation and the categorical body size measures among food insecure boys is that the sample size was too small to produce statistically reliable (precise) odds ratio estimates. There were only 22 observations in the referent category and just a fraction of these would have had the body size outcome of interest. For the remaining items in Part I below, limit the analyses to food secure boys. 2. Examine outcome: BMI z-score a. Calculate the following descriptive statistics. Please be aware of the footnotes to the table below. When using decimals report them to the nearest hundredth. Number 439 Mean* 0.60 SE* 0.09 Min -3.39 Max 3.33 Median** 0.56 Skewness (p-value)**,† 0.07 Kurtosis (p-value)**, † 0.02 *From Stata svy command; SAS users use design variables as you learned in E.6. **Stata users: Include aweight †SAS users: skip skewness and kurtosis R users: may be able to get Skewness and Kurtosis coefficients but not p-values . Peer-reviewers: please do not mark this problem wrong if the student is using R. b. Create a histogram and box plot. Paste below. Reduce the size to about 2”x2” if possible. In your project analysis, you should also use other normality plots (e.g., Q-Q) as appropriate for your study. Survey commands for these plots are not available, so they run as described in your Biostatistics courses. R users can us e svyhist() and svyboxplot() to include survey weights in this question. If so, their plots will look slightly different from STATA plots.

0-100% 209 101-200% 230 Health Insurance (hinsur) Any private 99 Public only 254 Other 86 i. Do the cell counts for these variables seem to be of sufficient size for analysis? Yes, the cell counts for both of these variables are of sufficient size for analysis. Part II: Descriptive Assessments: Bivariable Analyses For the questions in this section, limit the analyses to food secure boys . 1. Cross-tabulate categorical independent variables with a dependent categorical variable, high waist circumference. Outcome (counts, unweighted) Independent variables ≤ recommended waist circumference > recommended waist circumference Exposure: Food Assistance Yes 289 60 No 74 16 Potential Confounding Variable: Health Insurance (hinsur) Any private 82 17 Public only 213 41 Other 68 18 a. Are the cell counts for this categorical outcome variable of sufficient size for analysis? Do these cross-tabulations raise any concerns for logistic regression analysis with regard to statistical precision of odds ratio estimates? The cross-tabulation demonstrates that there are cell counts for the category, waist circumference larger than recommended (high waist circumference), that may not of sufficient size for analysis. Among food secure boys not receiving food assistance and among those with ‘Other’ health insurance type, fewer than 20 have high waist circumference. These counts raise concern about statistical precision of the associations involving this outcome variable. 2. Calculate 10 th and 90 th percentile values of continuous independent variables within each category of the dependent variable (high waist circumference). Outcome (counts, unweighted) ≤ recommended waist circumference > recommended waist circumference Independent variables 10 th percentile 90 th percentile 10 th percentile 90 th percentile Age ( ridageyr) 4 15 5 14 Stata users: Do not include aweights in the percentile calculations

- In this analysis, where we have < 2,000 values, use the Shapiro-Wilk test */ /* Note that this is not a survey procedure, means will not be accurate */ proc univariate data = ex7 normal ; *specify that we are testing for normal distribution; var bmiz; qqplot bmiz / Normal ( mu =est sigma =est color =red l = 1 ); *code to test for normality and to generate qqplots (not shown in video); histogram / normal ; *code to create histogram; where (subpop2 = 1 and male = 0 and foodinsec = 0 ); *weight wtmec2yr; *commented out because no weight variables when looking at normality tests in SAS; run ; Refer to the E7 VideoCode file for creation of the box plot. There are two options, that you can try: pro c x boxplot and proc sgplot. R Sample code 2a . **N, min, and max do not need to be survey weighted** ```{r} library( pastecs) round( stat.desc(bmiz[subpop2==1 & male=='boys' & foodinsec==0], desc=FALSE), 2) # all you need here is nbr.val, min, & max ``` ```{r} svymean (~bmiz, design=subset(subpop, male=="boys" & foodinsec==0)) #getting the median (choosing quantile = 0.5 for median only) svyquantile (~bmiz, design=subset( subpop, male=="boys" & foodinsec==0), quantile=0.5) #skew and kurtosis ##I do not know how to apply survey weights to skew or kurtosis in R for p-value, this is the closest I can figure out library( DescTools) Skew( bmiz[subpop2==1 & male=='boys' & foodinsec==0], weights = wtmec2 yr[subpop2==1 & male=='boys' & foodinsec==0]) Kurt( bmiz[subpop2==1 & male=='boys' & foodinsec==0], weights = wtmec2 yr[subpop2==1 & male=='boys' & foodinsec==0]) ``` 2b. Create a histogram and box plot. In R, you can use the survey weights to do this. R outputs will look slightly different from STATA because of this ```{r}