91Ó°ÊÓ

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,

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

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.

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

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

$$\begin{gathered} Z_{\frac{\mathrm{Î\pm }}{2}}=Z_{0.025} \\ =1.96 \end{gathered}$$

The formula for sample size is given below:

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

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