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In 2011, the Institute of Medicine (IOM), a non-profit group affiliated with the US National Academy of Sciences, reviewed a study measuring bone quality and levels of vitamin-D in a random sample from bodies of 675 people who died in good health. 8.5% of the 82 bodies with low vitamin-D levels (below 50 nmol/L) had weak bones. Comparatively, 1% of the 593 bodies with regular vitamin-D levels had weak bones. Is a normal model a good fit for the sampling distribution

User Mizbella
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Answer:

A normal model is a good fit for the sampling distribution.

Explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:


\mu_(\hat p)=p

The standard deviation of this sampling distribution of sample proportion is:


\sigma_(\hat p)=\sqrt{(p(1-p))/(n)}

The information provided is:

N = 675

X₁ = bodies with low vitamin-D levels had weak bones

n₁ = 82

p₁ = 0.085

X₂ = bodies with regular vitamin-D levels had weak bones

n₂ = 593

p₂ = 0.01

Both the sample sizes are large enough, i.e. n₁ = 82 > 30 and n₂ = 593 > 30.

So, the central limit theorem can be applied to approximate the sampling distribution of sample proportions by the Normal distribution.

Thus, a normal model is a good fit for the sampling distribution.

User Nnm
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