Step 1 It is given that 51 out of 298 orders are not accurate. The sample proportion is \(\displaystyle\frac{{51}}{{298}}={0.17}\). The sample size is 298. For the confidence interval 90%, the two-tailed z value at 0.05 level of significance is 1.645. The 90% confidence interval for the population proportion is calculated as follows: \(\displaystyle{C}{I}=\hat{{{p}}}\pm{z}_{{\frac{\alpha}{{2}}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)

\(\displaystyle={0.171}\pm{\left({1.645}\right)}\sqrt{{{\frac{{{0.17}{\left({1}-{0.17}\right)}}}{{{298}}}}}}\)

\(\displaystyle={0.171}\pm{0.036}\)

\(\displaystyle={\left({0.135},{0.207}\right)}\) Thus, the percentage of orders that is not accurate of restaurant A lies between 0.135 and 0.207. Step 2 The percentage of orders that are not accurate of restaurant B lies between 0.127 and 0.191. From the intervals it can be observed that neither restaurant appears to have a significant different percentage of orders that are not accurate because these two confidence intervals are overlapped.