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.

Short answer

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

Step-by-step solution

Given information

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

Concept

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

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 $12.284ft$