Chapter 6: Q50 (page 357)

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

Expert verified

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:

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

Compute the confidence interval of p

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

$(1 - α) = 0.95 α = 0.05$

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

$Z_{\frac{α}{2}} = Z_{0.025} = 1.96$

The 95% confidence interval for the proportion is computed as

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

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.