Q 8.78 - Elementary Statistics

Question

American Alligators. Multi-sensor data loggers were attached to free-ranging American alligators in a study conducted by Y. Watanabe for the article "Behavior of American Alligators Monitored by Multi-Sensor Data Loggers" (Aquatic Biology, Vol. 18 , pp. 1-8). The mean duration for a sample of $68$ dives was $338.0$ seconds. Assume the population standard deviation is $100$ seconds.

a. Find a $95\%$ confidence interval for the mean duration, $\mu$ of an American-alligator dive.

b. Repeal part (a) at a $99\%$ confidence level.

c. Draw a graph similar to Fig. $8.6$ on page 326 to display both confidence intervals.

d. Which confidence interval yields a more accurate estimate of $\mu$ ? Explain your answer.

Step-by-Step Solution

Answer

a. The confidence interval for $95\%$ confidence interval of $\mu$ is $(314.23, 361.77)$.

b. The confidence interval for $99\%$ confidence interval of $\mu$ is (306.77,369.23).

c. The graph is 

d. The $99\%$ confidence level of $\mu$ measures more accurate estimate.

Step 1: Given Information

Confidence level of $95\%, 99\%$ are given.

$\overline{x}=338, n=68.$

Step 2: Determining 95 % confidence level of μ

It is given that

$\overline{x}=338, n=68$

$95\%$ confidence interval is

$\overline{x}\pm z_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)=333\pm 1.96\left(\frac{10\pi }{\sqrt{68}}\right)$

$=338\pm 23.7685$

$=(314.23,361.77)$

Step 3: Computing confidence interval for 99 % for μ

It is given by

$\overline{x}\pm z_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{k}}\right)=338\pm 2575\left(\frac{100}{\sqrt{68}}\right)$

$=338\pm 312264$

$=(306.77, 369.23)$

Step 4: Plotting of graph

For $95\%$ of $\mu$

For $99\%$ of $\mu$

Step 5: Explanation of which confidence interval yields a more accurate estimate of μ

$95\%$ confidence interval of $\mu$ measure more accurate estimate than the other because of it shorter interval window.