Question: American Alligators. Refer to Exercise a. Determine the margin of erro
ID: 3132551 • Letter: Q
Question
Question: American Alligators. Refer to Exercise a. Determine the margin of error for the 95% confidence interval. b. Determine the margin of error for the 99% confidence interval c. Compare the margins of error found in parts (a) and (b). d. What principle is being illustrated? Exercise 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.Explanation / Answer
a)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 100
n = sample size = 68
Thus,
Margin of Error E = 23.7680545 [ANSWER]
******************
b)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
z(alpha/2) = critical z for the confidence interval = 2.575829304
s = sample standard deviation = 100
n = sample size = 68
Thus,
Margin of Error E = 31.23651851 [ANSWER]
*********************
c)
The margin of error of the 99% confidence interval was larger.
**********************
d)
Larger confidence levels yield wider confidence intervals.
*******************************************
Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.