Question

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

Answer 1

Answer:

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

Step-by-step 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:

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

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.