Question

The ability of lizards to recognize their predators via tongue flicks can often mean life or death for lizards. Seventeen juvenile common lizards were exposed to the chemical cues of the viper snake. Their responses, in number of tongue flicks per 20 minutes, are presented below. 727,217, 268, 438, 625, 319, 200, 591, 574, 727, 693, 336, 302, 761, 268, 353, 370 Assuming a normal distribution, find a 90% confidence interval for the mean number of tongue flicks per 20 minutes for all juvenile common lizards. Confidence interval:

Answer 1

Answer:

Confidence level is (380.7133, 533.2867)

Step-by-step explanation:

Responses in number of tongue flicks per 20 minutes of lizards, are: 727,217, 268, 438, 625, 319, 200, 591, 574, 727, 693, 336, 302, 761, 268, 353, 370

n = 17

Mean (μ) is:

$\mu=\frac{{\Sigma}x}{n} = \frac{727+217+ 268+ 438+ 625+ 319+ 200+ 591+ 574+ 727+ 693+ 336+ 302+ 761+ 268+ 353+ 370}{17} = 439.3529$

Standard deviation (σ) is:

$\sigma=\sqrt{\frac{\Sigma(x-\mu^2)}{n} } =\sqrt{\frac{(727-457)^2+(217-457)^2+...+(370-457)^2}{17} } =191.2$

The confidence interval (c) = 90% = 0.9

$\alpha=1-0.9=0.1\\\frac{\alpha }{2} = 0.05\\Z_{\frac{\alpha }{2} }=1.64$

Margin of error (e) = $=Z_{\frac{\alpha}{2} }*\frac{\sigma}{\sqrt{n} }=1.64*\frac{191.2}{\sqrt{17} }=76.2867$

Confidence level = μ ± e = 457 ± 76.2867 = (380.7133, 533.2867)