Question

Traffic sign maintenance. Refer to the Journal of Transportation Engineering (June 2013) study of traffic sign maintenance in North Carolina, Exercise 8.54 (p. 489). Recall that the proportion of signs on NCDOT-maintained roads that fail minimum requirements was compared to the corresponding proportion for signs on county-owned roads. How many signs should be sampled from each maintainer to estimate the difference between the proportions to within .03 using a 90% confidence interval? Assume the same number of signs will be sampled from NCDOT-maintained roads and county-owned roads

Short answer

The required sample size is 1504.

Step-by-step solution

Given Information

The confidence coefficient is 0.90

The sampling error is

$SE=.03$

Z-value

For $\alpha =0.1$

The z-value is given by

$$\begin{gathered} z_{\frac{\alpha }{2}}=z_{0.05} \\ =1.645 \end{gathered}$$

Compute the sample

For, $z=1.645,p_1=p_2=0.5 andSE=.03$

The sample is calculated as

$$\begin{gathered} n_1=n_2 \\ =\frac{{\left(z_{0.05}\right)}^2\times \left[p_1\left(1-p_1\right)+p_2\left(1-p_2\right)\right]}{{\left(SE\right)}^2} \\ =\frac{{1.645}^2\times \left[0.25+0.25\right]}{{0.03}^2} \\ =\frac{2.706025\times .5}{0.0009} \\ =\frac{1.3530}{0.0009} \\ =1503.33 \\ \approx 1504 \end{gathered}$$

Therefore, the required sample size is 1504.