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Smelling Out the Enemy. Snakes deposit chemical trails as they travel through their habitats. These trails are often detected and recognized by lizards, which are potential prey. The ability to recognize their predators via tongue flicks can often mean life or death for lizards. Scientists from the University of Antwerp were interested in quantifying the responses of juveniles of the common lizard to natural predator cues to determine whether the behavior is learned or congenital. Seventeen juvenile common lizards were exposed to the chemical cues of the viper snake. Their responses, in number of tongue flicks per twenty minutes, are presented in the following

Find and interpret a $90\%$confidence interval for the mean number of tongue flicks per $20$minutes for all juvenile common lizards. Assume a population standard deviation of $190.0$.

Step by step solution

01

Given Information

The response of common lizards in number of flicks per$20$ minutes is given below:

Also, population standard deviation is$190$.

02

Determination of sample mean

It is given that

$\underset{i=1}{\overset{13}{∑}}{x}_i=9047$

Therefore, sample mean$\stackrel{¯}{x}=\frac{1}{n}×\underset{i=1}{\overset{n}{∑}}xi=\frac{1}{17}×9047=532.1765$

03

Determination of confidence interval

For $90\%$confidence interval of $\mathrm{μ}$,

confidence level $=90\%=100×0.90\%$

$1-\mathrm{Î\pm }=0.90$

$⇒\mathrm{Î\pm }=0.10$

Hence, confidence interval of $\mathrm{μ}$is

$\stackrel{¯}{x}-z_{0.05}×\frac{\mathrm{σ}}{\sqrt{n}},\stackrel{¯}{x}+z_{0.05}×\frac{\mathrm{σ}}{\sqrt{n}}$

$=532.1765-1.645×\frac{190}{\sqrt{17}},532.1765+1.645×\frac{190}{\sqrt{17}}$

$=532,1765-75.8045,532,1765+75.8045$

$=456.372,607.9810$

Therefore,$90\%$confidence interval of$\mathrm{μ}$is$\left(456.372,607.9810\right)$

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