#### Answer

a) p is between 0.5596 and 0.6204.
b) Could be biased.
c)1083

#### Work Step by Step

a)$\alpha=1-0.95=0.05.$ $\sigma$ is known, hence we use the z-distribution. $z_{\alpha/2}=z_{0.025}=1.96.$ Margin of error:$E=z_{\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat{p}\cdot (1-\hat{p})}{n}}=1.96\cdot \sqrt{\frac{0.59\cdot (1-0.59)}{1003}}=0.0304.$ Hence the confidence interval: p is between 0.59-0.0304=0.5596 and 0.59-0.0304=0.6204.
b)The survey questions involve hand sanitation and the survey was sponsored by a hand sanitizer producer company, hence the results could be biased for the company to benefit from it.
c)$\alpha=1-0.9=0.1.$ The corresponding z-value: $z_{\alpha/2}=z_{0.05}=1.645.$ $\hat{p}$ is unknown, hence the sample size: $n=\frac{z^2_{\frac{\alpha}{2}}\cdot0.25}{E^2}=\frac{1.645^2 \cdot 0.25}{0.025^2}\approx1083.$