Q 8.80 - Elementary Statistics

Question

American Alligators. Refer to Exercise 8.78.

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?

Step-by-Step Solution

Answer

(a) The margin of error for $95\%$ confidence interval is $23.77$ seconds.

(b) The margin of error for $99\%$ confidence interval is $31.23$ seconds.

(c) Margin of error of $95\%$confidence level is smaller than that of $99\%$ confidence level.

(d) Same sample size and increased confidence level.

Step 1: Given Information

From previous exercise, $95\%$ confidence level of $\mu$ is $(314.23,361.77)$.

$99\%$ confidence level of $\mu$ is $(306.77,369.23)$

Step 2: Margin of Error for 95 % Confidence Interval

Margin of error is calculated as:

$E=\frac{\mathrm{Upper} \mathrm{limit}-\mathrm{Lower} \mathrm{limit}}{2}$

$E=\frac{361.77-314.23}{2}$

$E=\frac{47.54}{2}$

$E=23.77$seconds

Step 3: Margin of Error for 99 % Confidence Interval

Margin of Error is given by

$\text{ Margin of error }=\frac{\text{ Upper limit }-\text{ Lower limit }}{2}$

$E=\frac{369.23-306.77}{2}$

$E=\frac{62.46}{2}$

$E=31.23$seconds

Step 4: Comparison of the margins of error found above

It is found from above results that confidence interval of $95\%$ has smaller margin of error than that of $99\%$ confidence interval.

Step 5: Principle Illustrated

The principle illustrated is:

When sample size is same and confidence level is increased, margin of error increases.