College

A similar survey resulted in a sample mean of [tex]\$1.69[/tex] and 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]

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

Answer :

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

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

2. Determine the Z-Score for the Confidence Level:
For an 80% confidence level, the area in each tail of the normal distribution is 10% (since 80% is in the middle). You can find the Z-score corresponding to this by looking up the value that accumulates 90% of the distribution in Z-score tables or using a calculator. The Z-score for an 80% confidence interval is approximately 1.282.

3. Calculate the Standard Error (SE):
The standard error of the mean is calculated with the formula:
\[
SE = \frac{\sigma}{\sqrt{n}}
\]
Substituting the given values:
\[
SE = \frac{0.657}{\sqrt{50}} \approx 0.0929
\]

4. Calculate the Margin of Error (ME):
The margin of error is found by multiplying the Z-score by the standard error:
\[
ME = Z \times SE = 1.282 \times 0.0929 \approx 0.119
\]

5. Determine the Confidence Interval:
The confidence interval is then calculated as:
\[
\text{Confidence Interval} = \left(\bar{x} - ME, \bar{x} + ME\right)
\]
So:
\[
= (1.69 - 0.119, 1.69 + 0.119) = (1.571, 1.809)
\]

Therefore, the 80% confidence interval for the population mean is approximately \$[/tex]1.69 ± \$0.119. The correct answer from the choices given is [tex]\(1.69 \pm 0.119\)[/tex].

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