In a similar survey, a sample mean of [tex]\$1.69[/tex] was obtained with a population standard deviation of 0.657. If 50 people completed the survey, what is the [tex]80\%[/tex] confidence interval for the population mean?

A. [tex]1.69 \pm 0.156[/tex]
B. [tex]1.69 \pm 0.182[/tex]
C. [tex]1.69 \pm 0.119[/tex]
D. [tex]1.69 \pm 0.657[/tex]

(Note: If necessary, use the Standard Normal Distribution Z-Score Table.)

To find the 80% confidence interval for the population mean, we'll follow these steps:

1. Identify the given information:
- Sample mean ([tex]$\bar{x}$[/tex]) = [tex]$1.69
- Population standard deviation ($\sigma$) = $[/tex]0.657
- Sample size ([tex]$n$[/tex]) = 50
- Confidence level = 80%

2. Find the Z-score for the 80% confidence level:
The 80% confidence level corresponds to a Z-score that leaves 10% in each tail of the normal distribution. We use the standard normal distribution table and find the Z-score for 40% in the tail (since 80% confidence means 20% total in the tails, split in two).

3. Calculate the standard error of the mean (SEM):
[tex]\[
\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{0.657}{\sqrt{50}}
\][/tex]

4. Calculate the margin of error:
[tex]\[
\text{Margin of Error} = Z \times \text{Standard Error}
\][/tex]
- Using the Z-score for the 80% confidence level (which is approximately 1.282), calculate:
[tex]\[
\text{Margin of Error} = 1.282 \times \frac{0.657}{\sqrt{50}}
\][/tex]

5. Determine the 80% confidence interval:
- Lower bound: [tex]$\bar{x} - \text{Margin of Error} = 1.69 - 0.119$[/tex]
- Upper bound: [tex]$\bar{x} + \text{Margin of Error} = 1.69 + 0.119$[/tex]

Using these calculations, the 80% confidence interval for the population mean is approximately:
[tex]\[
1.69 \pm 0.119
\][/tex]

Therefore, the correct answer is [tex]$1.69 \pm 0.119$[/tex].