Question: Eye shadow, mascara, and nickel allergies. Refer to the Journal of the European Academy of Dermatology and Venereology (June 2010) study of the link between nickel allergies and use of mascara or eye shadow, Exercise 6.58 (p. 358). Recall that two groups of women were a sampled-one group with cosmetic dermatitis from using eye shadow and another group with cosmetic dermatitis from using mascara. In either group, how many women would need to be sampled in order to yield an estimate of the population percentage with a nickel allergy that falls no more than 3% from the true value?

From Exercise 6.58,

The number of women in the first sample=131.

The number of women in the second sample=250.

The number of women diagnosed with a nickel allergy while using the eye shadow in the first sample=12.

The number of women diagnosed with a nickel allergy while using the mascara in the second sample=25.

The sample fraction of success for the first group with cosmetic dermatitis from using eye shadow is,

$$\begin{gathered} \hat{p}=\frac{12}{131} \\ =0.09 \\ \approx 0.1 \end{gathered}$$

The sample fraction of success for the second group with cosmetic dermatitis from using mascara is,

$$\begin{gathered} \hat{p}=\frac{25}{250} \\ =0.1 \end{gathered}$$

Let the confidence level be 0.95.

$$\begin{gathered} For\left(1-\alpha \right)=0.95 \\ \alpha =0.05 \\ \frac{\alpha }{2}=0.025 \end{gathered}$$

The$Z_{\frac{\alpha }{2}}$corresponding to the standard normal table is,

$$\begin{gathered} Z_{\frac{\alpha }{2}}=Z_{0.025} \\ =1.96 \end{gathered}$$

The formula for sample size is given below:

$n=\frac{{\left(Z_{\frac{\alpha }{2}}\right)}^2\left(pq\right)}{{\left(SE\right)}^2}$

Where SE is the sampling error.

The value of the pq is unknown; it can be estimated by using the sample fraction of success, from a prior sample

Let the sample proportion $\hat{p}$of both groups be near 0.1.

Here, the value pq is unknown. Which can be obtained by using the sample fraction of success$\hat{p}$

The product of pq is computed as:

$$\begin{gathered} pq=p\left(1-p\right) \\ =\left(0.1\right)\left(1-0.1\right) \\ =\left(0.1\right)\left(0.9\right) \\ =0.09 \end{gathered}$$

The sampling error is 3%.

i.e$$\begin{gathered} SE=\frac{3}{100} \\ =0.03 \end{gathered}$$

The sample size is computed as