Q 8.134. - Elementary Statistics

Question

Bottlenose Dolphins. The webpage "Bottlenose Dolphin" produced by the National Geographic Society provides information about the bottlenose dolphin. A random sample of $50$adult bottlenose dolphins have a mean length of $12.04\mathrm{ft}$ with a standard deviation of $1.03\mathrm{ft}$ Find and interpret a $90\%$ confidence interval for the mean length of all adult bottlenose dolphins.

Step-by-Step Solution

Answer

The $90\%$confidence interval for $\mu$ is $(11.796,12.284)$

Step 1: Given information

The average length of $50$ adult bottlenose dolphins was $12.04\mathrm{ft}$ with a standard deviation of $1.03\mathrm{ft}$

Step 2: Concept

The formula used:  $Z=\overline{x}\pm t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)$

Step 3: Calculation

Find a $90\%$ confidence interval for all adult bottlenose dolphins' mean length. 
Consider $\overline{x}=12.04, n=50, s=1.03$, and the confidence level is $90\%$

From "Table IV Values of $t_{\alpha }$ " the required value  $t_{\frac{\alpha }{2}}$ for $90\%$ confidence with $49 (=50-1)$ degrees of freedom is $1.677$

The $90\%$ confidence interval is,

$$\begin{gathered} \overline{x}\pm t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)=12.04\pm 1.677\left(\frac{1.03}{\sqrt{50}}\right) \\ =12.04\pm 0.2443 \\ =(55-3.3688,55+3.3688) \\ =(11.796,12.284) \end{gathered}$$

As a result, $(11.796,12.284)$ is the $90\%$ confidence interval for $\mu$

The $90\%$ confidence interval for all adult bottlenose dolphins' mean length is between$11.796\mathrm{ft}$ and $12.284ft$