Compute the standard error for sample means from a population with mean $$ \mu=120 $$ and standard deviation $$ \sigma=40 $$ for sample sizes of $$ n=45, n=230 $$, and $$ n=1100 $$. Round your answers to two decimal places. Sample Size $$ n=45: $$ Standard Error $$ n=230: $$ i $$ n=1100: $$ i

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**Step 1: Write down the formula for the standard error (SE).**  
The standard error is given by  
$$
\text{SE} = \frac{\sigma}{\sqrt{n}}
$$

**Step 2: Plug in the values for $$ n=45 $$.**  
With $$\sigma=40$$ and $$ n=45 $$:  
$$
\text{SE} = \frac{40}{\sqrt{45}}
$$  
First, calculate $$\sqrt{45}$$:  
$$
\sqrt{45} \approx 6.71
$$  
Then,  
$$
\text{SE} \approx \frac{40}{6.71} \approx 5.96
$$

**Step 3: Plug in the values for $$ n=230 $$.**  
With $$\sigma=40$$ and $$ n=230 $$:  
$$
\text{SE} = \frac{40}{\sqrt{230}}
$$  
Calculate $$\sqrt{230}$$:  
$$
\sqrt{230} \approx 15.17
$$  
Then,  
$$
\text{SE} \approx \frac{40}{15.17} \approx 2.64
$$

**Step 4: Plug in the values for $$ n=1100 $$.**  
With $$\sigma=40$$ and $$ n=1100 $$:  
$$
\text{SE} = \frac{40}{\sqrt{1100}}
$$  
Calculate $$\sqrt{1100}$$:  
$$
\sqrt{1100} \approx 33.17
$$  
Then,  
$$
\text{SE} \approx \frac{40}{33.17} \approx 1.21
$$

**Final Answers:**  
- Sample Size $$ n=45 $$: Standard Error $$\approx 5.96$$  
- Sample Size $$ n=230 $$: Standard Error $$\approx 2.64$$  
- Sample Size $$ n=1100 $$: Standard Error $$\approx 1.21$$
- For $$ n=45 $$, the standard error is approximately 5.96. - For $$ n=230 $$, the standard error is approximately 2.64. - For $$ n=1100 $$, the standard error is approximately 1.21.

Compute the standard error for sample means from a population with mean \( \mu=120 \) and standard deviation \( \sigma=40 \) for sample sizes of \( n=45, n=230 \), and \( n=1100 \). Round your answers to two decimal places. Sample Size \( n=45: \) Standard Error \( n=230: \) i \( n=1100: \) i

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