SOLUTION: 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

Question 1180147: 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. Scientist from Kenyatta University were interested in quantifying the responses of Juveniles of the common lizard to the 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 20 minutes, are presented in the following table:
425 510 629 236 654 200 710 276 633
501 811 332 424 674 676 662 694
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 190.
Interpret your answer from part (i).

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to calculate the confidence interval and interpret it:
**1. Calculate the Sample Mean (x̄):**
Add all the tongue flick counts and divide by the number of lizards (17):
x̄ = (425 + 510 + 629 + 236 + 654 + 200 + 710 + 276 + 633 + 501 + 811 + 332 + 424 + 674 + 676 + 662 + 694) / 17
x̄ = 8741 / 17
x̄ ≈ 514.18 tongue flicks
**2. Identify Given Information:**
* Sample size (n) = 17
* Population standard deviation (σ) = 190
* Confidence level = 90%
**3. Find the Critical Z-Value:**
For a 90% confidence level, α = 1 - 0.90 = 0.10. α/2 = 0.05. We need to find the z-score that corresponds to an area of 0.95 (1 - 0.05) in the standard normal distribution. This z-value is approximately 1.645.
**4. Calculate the Margin of Error (E):**
E = z * (σ / √n)
E = 1.645 * (190 / √17)
E ≈ 1.645 * (190 / 4.123)
E ≈ 1.645 * 46.08
E ≈ 75.71 tongue flicks
**5. Construct the Confidence Interval:**
Lower Bound = x̄ - E = 514.18 - 75.71 ≈ 438.47 tongue flicks
Upper Bound = x̄ + E = 514.18 + 75.71 ≈ 589.89 tongue flicks
Therefore, the 90% confidence interval is approximately (438.47, 589.89).
**Interpretation:**
We are 90% confident that the true mean number of tongue flicks per 20 minutes for all juvenile common lizards when exposed to viper snake chemical cues is between 438.47 and 589.89 tongue flicks. This means that if we were to repeat this study many times and calculate a 90% confidence interval each time, 90% of those intervals would contain the true population mean.

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