 # Comparison between two diets in obese children, one low-fat and one low-carb 8 week intervention with the outcome of weight loss. Based on a similar study in adults, 20% of the children will not complete the study. For a 95% confidence interval with a margin of error of no more than 3 lbs., how many children should be recruited? In the adult trial, the low-fat and low-carb groups had a standard deviations of 8.4 and 7.7, respectively and each group had 100 participants. A. Calculate the pooled standard deviation (Sp). B. Calculate the number of children will need to be recruited for the trial if all finish the study use the Sp calculated for the standard deviation? C. How many children would be needed to account for 20% attrition. pliwraih 2022-07-22 Answered
Comparison between two diets in obese children, one low-fat and one low-carb 8 week intervention with the outcome of weight loss. Based on a similar study in adults, 20% of the children will not complete the study. For a 95% confidence interval with a margin of error of no more than 3 lbs., how many children should be recruited? In the adult trial, the low-fat and low-carb groups had a standard deviations of 8.4 and 7.7, respectively and each group had 100 participants.
A. Calculate the pooled standard deviation (Sp).
B. Calculate the number of children will need to be recruited for the trial if all finish the study use the Sp calculated for the standard deviation?
C. How many children would be needed to account for 20% attrition.
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A).
The standard deviations of low-fat and low-carb groups are 8.4 and 7.7 respectively. The sample size of both the groups is 100.
The pooled standard deviation can be calculated as follows:
${s}_{p}=\frac{\left(n-1\right){s}_{1}^{2}+\left({n}_{2}-1\right){s}_{2}^{2}}{{n}_{1}+{n}_{2}-2}\phantom{\rule{0ex}{0ex}}=\frac{\left(100-1\right){8.4}^{2}+\left(100-1\right){7.7}^{2}}{100+100-2}\phantom{\rule{0ex}{0ex}}\approx 8.05$
B).
The pooled standard deviation = 7.97
t-value for 95% confidence interval at 198 df is 1.97.
Sample size is 100 for both the samples.
The margin of error is 3.
$\text{Margin of error}=t\left(\frac{{s}_{p}}{\sqrt{n}}\right)\phantom{\rule{0ex}{0ex}}3=1.97\left(\frac{8.05}{\sqrt{n}}\right)\phantom{\rule{0ex}{0ex}}\sqrt{n}=5.286\phantom{\rule{0ex}{0ex}}n=27.94$
Therefore, the number of children needed is 30.
C).
It is given that the proportion of attrition is 0.20
$\text{Margin of error}=t\left(\sqrt{\frac{p\left(1-p\right)}{n}}\right)\phantom{\rule{0ex}{0ex}}3=1.97\left(\sqrt{\frac{0.2\left(1-0.2\right)}{n}}\right)\phantom{\rule{0ex}{0ex}}\sqrt{n}=0.2627\phantom{\rule{0ex}{0ex}}n=0.07$
Therefore, the number of children to account for 20% attrition is 1.
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