In a grocery store survey, a sample mean of [tex]\$1.69[/tex] and a population standard deviation of 0.657 were observed. If 50 people completed the survey, what is the [tex]80\%[/tex] confidence interval for the population mean?

If necessary, use the Standard Normal Distribution Z-Score Table.

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]

To calculate the 80% confidence interval for the population mean based on the given survey data, 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 Z-score represents the number of standard deviations a data point is from the mean. For an 80% confidence level, the Z-score is approximately 1.2816. This value is typically found using a standard normal distribution Z-score table.

3. Calculate the standard error:
The standard error of the mean is calculated using the formula:

[tex]\[
\text{Standard Error} (SE) = \frac{\sigma}{\sqrt{n}}
\][/tex]

Substituting the values:

[tex]\[
SE = \frac{0.657}{\sqrt{50}} \approx 0.0929
\][/tex]

4. Calculate the margin of error:
The margin of error is calculated by multiplying the Z-score by the standard error:

[tex]\[
\text{Margin of Error} = Z \times SE = 1.2816 \times 0.0929 \approx 0.1191
\][/tex]

5. Determine the confidence interval:
The confidence interval is calculated by adding and subtracting the margin of error from the sample mean:

[tex]\[
\text{Lower limit} = \bar{x} - \text{Margin of Error} = 1.69 - 0.1191 \approx 1.5709
\][/tex]

[tex]\[
\text{Upper limit} = \bar{x} + \text{Margin of Error} = 1.69 + 0.1191 \approx 1.8091
\][/tex]

The 80% confidence interval for the population mean is approximately [tex]$1.69 \pm 0.119$[/tex]. This matches the answer choice: [tex]$1.69 \pm 0.119$[/tex].