Q 8.82. - Elementary Statistics

Question

American Alligators. Refer to Exercise $8.78$

a. The mean duration for a sample of $612$ dives was $322$ seconds. Find a $995$ confidence interval for $\mu$ based on that data.

b. Compare the $995$ confidence intervals obtained here and in Exercise $8.78$ (b) by drawing a graph similar to Fig. $8.7$ on page $327$

c. Compare the margins of error for the two $99\%$ confidence intervals.

d. What principle is being illustrated?

Step-by-Step Solution

Answer

Part (a) $(311.588, 332.412)$

Part (b)  

Part (c) $31.23$

Part (d) The sample size is increased and the confidence level is the same, which provides a decreasing margin of error.

Part (a) Step 1: Given information

$\overline{x}=322, n=612$ and $\sigma =100$

Part (a) Step 2: Concept

The formula used:  Margin of error.

Part (a) Step 3: Calculation

Using MINITAB, compute a $99\%$ confidence interval estimate of the population mean $\mu$

Consider $\overline{x}=322, n=612$ and $\sigma =100$

Procedure for MINITAB: 

Step 1: Select Stat $>$ Basic Statistics $>$ $1-$Sample $Z$ from the drop-down menu.

Step 2: In the Summarized Data section, enter $612$ as the sample size and $322$ as the mean.

Step 3: In the Standard deviation box, type $100$ for $s$

Step 4: Select Options and enter $322$ as the level of confidence.

Step 5: In the alternative, select not equal.

Step 6: In all dialogue boxes, click OK. 

MINITAB output:

One-Sample $Z$

The assumed standard deviation $=100$

N	Mean	SE Mean	99% CI
612	322.000	4.042	(311.588, 332.412)

The population mean $\mu$ has a $99\%$ confidence interval estimate of $(311.588,332.412)$ based on MINITAB output.

Part (b) Step 1: Explanation

 The confidence interval in Exercise $8.78$ is shown in the below graph: 

The confidence interval in part (a) is shown in the below graph: 

Part (c) Step 1: Calculation

When $(311.588,332.412)$ is used, get the margin of error for the $99\%$ confidence interval?

The margin of error is,

$$\begin{gathered} \text{ Margin of error }=\frac{332.412-311.588}{2} \\ =\frac{20.824}{2} \\ =10.412 \end{gathered}$$

Thus, the margin of error for the $99\%$ confidence interval is $10.412$

The $99\%$ confidence interval for $\mu$ is $(306.77,369.23)$ based on Exercise $8.78$

The margin of error is,

$$\begin{gathered} \text{ Margin of error }=\frac{369.23-306.77}{2} \\ =\frac{62.46}{2} \\ =31.23 \end{gathered}$$

Thus, the margin of error for the $\mu$ confidence interval is $31.23$

Comparison:

When compared to the margin of error determined in Exercise $8.78$ the margin of error for this exercise is less.

Part (d) Step 1: Explanation

The premise is that the sample size is raised while the confidence level remains constant, resulting in a smaller margin of error.