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.

540, 744, 710, 251, 285, 591, 472, 370, 268, 608, 421, 693, 506, 268, 200, 778, 574

Preliminary data analyses indicate that you can reasonably apply the z-interval procedure. Find 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. Note: The sum of the data is 8279.

Confidence interval: (____________,___________).

Answer 1

Answer:

$487-1.64\frac{14}{\sqrt{17}}=481.43$  

$487+1.64\frac{14}{\sqrt{17}}=492.57$  

So on this case the 90% confidence interval would be given by (481.43;492.57)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

We have the following dataset:

540, 744, 710, 251, 285, 591, 472, 370, 268, 608, 421, 693, 506, 268, 200, 778, 574

We can calculate the sample mean with the following formula:

$\bar x = \frac{\sum_{i=1}^n X_i}{n}$

$\bar X=487$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$\sigma=190$ represent the population standard deviation  

n=17 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)  

Develop a 90% confidence interval about the mean.

Since the confidence is 0.9 or 90%, the value of $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$  

Now we have everything in order to replace into formula (1):  

$487-1.64\frac{14}{\sqrt{17}}=481.43$  

$487+1.64\frac{14}{\sqrt{17}}=492.57$  

So on this case the 90% confidence interval would be given by (481.43;492.57)