Question

Smelling Out the Enemy. In Exercise $8.72$ you found a $90\%$ confidence interval for the mean number of tongue flicks per $20$ minutes for all juvenile common lizards to be from $456.4$ to $608.0$ Obtain the margin of error by

a. taking half the length of the confidence interval.

b. using Formula $8.1$on the page $325$(Recall that $n=17$and that$\sigma =190.0.)$

Short answer

Part (a)$75.8$

Part (b)$75.8045$

Step-by-step solution

Part (a) Step 1: Given information

For all juvenile common lizards, the $90\%\mathrm{Cl}$ for the mean number of tongue flicks every $20$ minutes ranges from $456.4$ to $608.0$

Part (a) Step 2: Concept

The formula used:$E=Z_{0.05}\times \frac{\sigma }{\sqrt{n}}$

Part (a) Step 3: Calculation

$\therefore$Length of the confidence interval

$$\begin{gathered} =608.0-456.4 \\ =151.6 \end{gathered}$$

$\therefore$The margin of error $=\frac{1}{2}\times$length of the confidence interval

$$\begin{gathered} =\frac{1}{2}\times 151.6 \\ =75.8 \end{gathered}$$

Part (b) Step 1: Calculation 

Population s.d., $\sigma =190$

Sample size $n=17$

Confidence level $=90\%$

$$\begin{gathered} =100\times 0.90\% \\ \therefore \alpha =1-0.90=0.1 \\ \therefore Z_{\frac{\alpha }{2}}=Z_{\frac{0.1}{2}}=Z_{0.05} \\ =1.645 \end{gathered}$$

$\therefore$The margin of error at $90\%$ confidence level

$$\begin{gathered} E=Z_{0.05}\times \frac{\sigma }{\sqrt{n}} \\ =\frac{1.645\times 190}{\sqrt{17}} \\ =75.8045 \end{gathered}$$

$\therefore$The margin of error is $75.8045$ tongue flicks$/20min$