Question

Question: Nannies who are INA certified. The International Nanny Association (INA) reports that in a sample of 928 in-home child care providers (nannies), 128 have passed the INA Nanny Credential Exam (2014 International Nanny Association Salary and Benefits Survey). Use Wilson’s adjustment to find a 95% confidence interval for the true proportion of all nannies who have passed the INA certification exam.

Short answer

We are 95% confident that the true proportion of all nannies that have passed the INA certification exam lies between the interval of 0.117 and 0.161.

Step-by-step solution

Given Information

The sample size was n=928.

The number of successes in the sample, x=128.

The confidence interval=95%.

Compute the adjusted sample proportion

The adjusted population proportion is calculated as:

$$\begin{gathered} \tilde{p}=\frac{x+2}{n+4} \\ =\frac{128+2}{928+4} \\ =0.139 \end{gathered}$$

Compute the confidence interval of p

The significance level corresponds to the 95% confidence interval is,

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

The value of is obtained from the standard normal table is,

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

The 95% confidence interval for the proportion is computed as

$$\begin{gathered} C.I=\left(\tilde{p}\pm Z_{\frac{\alpha }{2}}\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{n+4}}\right) \\ =\left(0.139\pm 1.96\times \sqrt{\frac{0.139\left(1-0.139\right)}{928+4}}\right) \\ =\left(0.139\pm 0.022\right) \\ =\left(0.117,0.161\right) \end{gathered}$$

Hence, we are 95% confident that the actual proportion of all nannies that have passed the INA certification exam lies between the interval of 0.117 and 0.161.