Question

Hi, can you please help me answer this question, thank you

Answer 1

Step 1

State the formula for the confidence interval

`\bar{x}\pm z_{(\alpha)/(2)}*(\frac{\sigma}{\sqrt[]{n}})`

Where;

`\begin{gathered} \bar{x}=mean=63\text{ ounces} \\ \sigma=\text{ standard deviation = 4 ounces} \\ n=\text{ 38 backpacks} \\ z_{(\alpha)/(2)}=z_(0.005)=2.576 \end{gathered}`

Step 2

Find the margin of error, E

`\begin{gathered} E=z_{(\alpha)/(2)}*(\frac{\sigma}{\sqrt[]{n}}) \\ E=2.576*\frac{4}{\sqrt[]{38}} \\ E=\text{ }1.671529523 \\ E\approx\text{ 1.67 to 2 decimal places} \end{gathered}`

Step 3

Find the final answer

`63\pm1.67\text{ ounces to 2 decimal places}`