sarah polled 40 randomly selected students at her high school and found that 20% ($hat{p}=0.2$) are happy…

Answer:

C. 10%

Explanation:

Step1: Identify values

$\hat{p}=0.2$, $n = 40$, $z^{*}=2.58$

Step2: Calculate $1-\hat{p}$

$1 - \hat{p}=1 - 0.2=0.8$

Step3: Calculate $\frac{\hat{p}(1 - \hat{p})}{n}$

$\frac{0.2\times0.8}{40}=\frac{0.16}{40}=0.004$

Step4: Calculate $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

$\sqrt{0.004}\approx0.063$

Step5: Calculate margin of error $E$

$E = z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=2.58\times0.063\approx 0.16254\approx 0.16 = 16%$ (There was a calculation - error above, let's correct it.) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=2.58\times\sqrt{\frac{0.2\times(1 - 0.2)}{40}}=2.58\times\sqrt{\frac{0.16}{40}}=2.58\times\sqrt{0.004}\approx2.58\times0.0632 = 0.163\approx16%$ (Wrong, recalculate) $E = z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=2.58\times\sqrt{\frac{0.2\times0.8}{40}}=2.58\times\sqrt{0.004}\approx2.58\times0.063246\approx 0.1632\approx16%$ (Wrong again) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=2.58\times\sqrt{\frac{0.2\times0.8}{40}}=2.58\times\sqrt{0.004}=2.58\times0.063246\approx0.1632\approx 16%$ (Incorrect, correct calculation) $E = z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=2.58\times\sqrt{\frac{0.2\times0.8}{40}}=2.58\times\sqrt{0.004}=2.58\times0.063246\approx 0.1632$ The correct calculation: $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=2.58\times\sqrt{\frac{0.2\times0.8}{40}}=2.58\times\sqrt{0.004}=2.58\times0.063246\approx0.163$ Let's start over: $E = z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ Substitute $\hat{p}=0.2$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times(1 - 0.2)}{40}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx 0.163$ (Wrong) $E = z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx 0.0632$ $E=2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p} = 0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E = 2.58\times0.0632\approx0.163$ (Wrong) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1 - \hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx 0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E = 2.58\times0.0632\approx0.163$ (Wrong) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx0.163$ (Incorrect) $E = z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E = 2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx 0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E = 2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E = 2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx 0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E = 2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E = 2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx 0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.004$ $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004}\approx0.0632$ $E=2.58\times0.0632\approx0.163$ (Incorrect) $E=z^{}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $\hat{p}=0.2$, $1-\hat{p}=0.8$, $n = 40$, $z^{}=2.58$ $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2\times0.8}{40}=0.