Solutions

Please note that the steps show rounded numbers, but that the final answers to the problems are calculated without rounding.

Problem Part Solution
1 - \(\hat{p} = \frac{x}{n}\)
x = number of successes
n = number of trials.
2 - \(\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)
3 - \(n \hat{p} \geq 10\)
\(n(1 - \hat{p}) \geq 10\)
4 - \(n = (\frac{z^*}{2m})^2\)
5 - \(z = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\)
6 - \(np \geq 10\)
\(n(1 - p) \geq 10\)
7 A \(\hat{p}=\frac{x}{n}=\frac{684}{1006}=0.6799\)
7 B \(n \hat{p} \geq 10\) -> \(1006* 0.6799\geq 10\) -> \(684 \geq 10\)
\(n(1 - \hat{p}) \geq 10\) -> \(1006(1-0.6799) \geq 10\) -> \(322 \geq 10\)
The requirements are met.
7 C \(1.96 * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 1.96 * \sqrt{\frac{0.6799(1-0.6799)}{1006}}=0.0288\)
7 D \(\hat{p} \pm \text{margin of error}\) -> \(0.6799 \pm 0.0288\) -> (0.6511, 0.7087)
We are 95% confident that the true population mean, p, is between 0.6511 and 0.7087.
7 E \(n = (\frac{z^*}{2m})^2 = (\frac{1.96}{2\times0.015})^2 = 4268.4444\) -> \(4269\)
8 A \(H_0: p = 0.6\)
\(H_a: p > 0.6\)
8 B \(z = \frac{0.6799-0.6}{\sqrt{\frac{0.6(1-0.6}{1006}}} = 5.174\)
8 C Applet
p = 0
8 D We reject the null hypothesis
8 E We have sufficient evidence to conclude that the true proportion of students that have been in a car accident within the last 5 years is greater than 0.6.