Answer :
To find the 80% confidence interval for the population mean, we use the formula
[tex]$$
\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}},
$$[/tex]
where
- [tex]$\bar{x}$[/tex] is the sample mean,
- [tex]$\sigma$[/tex] is the population standard deviation,
- [tex]$n$[/tex] is the sample size,
- [tex]$z_{\alpha/2}$[/tex] is the critical value from the standard normal distribution that leaves an area of [tex]$\alpha/2$[/tex] in the tail, and
- [tex]$\alpha = 1 - \text{confidence level}$[/tex].
Here, we have:
- Sample mean: [tex]$\bar{x} = 1.69$[/tex],
- Population standard deviation: [tex]$\sigma = 0.657$[/tex],
- Sample size: [tex]$n = 50$[/tex],
- Confidence level: [tex]$80\%$[/tex], so [tex]$\alpha = 0.20$[/tex]. This makes [tex]$\alpha/2 = 0.10$[/tex].
The critical value [tex]$z_{\alpha/2}$[/tex] corresponding to an upper-tail area of [tex]$0.10$[/tex] is approximately
[tex]$$
z_{0.10} \approx 1.28155.
$$[/tex]
Next, we calculate the standard error:
[tex]$$
\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{0.657}{\sqrt{50}}.
$$[/tex]
The value of [tex]$\sqrt{50}$[/tex] is about [tex]$7.071$[/tex], so
[tex]$$
\frac{0.657}{7.071} \approx 0.0929.
$$[/tex]
Now, compute the margin of error:
[tex]$$
\text{Margin of Error} = z_{\alpha/2} \cdot \text{Standard Error} = 1.28155 \times 0.0929 \approx 0.119.
$$[/tex]
Thus, the confidence interval is
[tex]$$
1.69 \pm 0.119.
$$[/tex]
In other words, the 80% confidence interval for the population mean is approximately
[tex]$$
(1.57, \, 1.81).
$$[/tex]
This matches the option [tex]$1.69 \pm 0.119$[/tex].
[tex]$$
\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}},
$$[/tex]
where
- [tex]$\bar{x}$[/tex] is the sample mean,
- [tex]$\sigma$[/tex] is the population standard deviation,
- [tex]$n$[/tex] is the sample size,
- [tex]$z_{\alpha/2}$[/tex] is the critical value from the standard normal distribution that leaves an area of [tex]$\alpha/2$[/tex] in the tail, and
- [tex]$\alpha = 1 - \text{confidence level}$[/tex].
Here, we have:
- Sample mean: [tex]$\bar{x} = 1.69$[/tex],
- Population standard deviation: [tex]$\sigma = 0.657$[/tex],
- Sample size: [tex]$n = 50$[/tex],
- Confidence level: [tex]$80\%$[/tex], so [tex]$\alpha = 0.20$[/tex]. This makes [tex]$\alpha/2 = 0.10$[/tex].
The critical value [tex]$z_{\alpha/2}$[/tex] corresponding to an upper-tail area of [tex]$0.10$[/tex] is approximately
[tex]$$
z_{0.10} \approx 1.28155.
$$[/tex]
Next, we calculate the standard error:
[tex]$$
\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{0.657}{\sqrt{50}}.
$$[/tex]
The value of [tex]$\sqrt{50}$[/tex] is about [tex]$7.071$[/tex], so
[tex]$$
\frac{0.657}{7.071} \approx 0.0929.
$$[/tex]
Now, compute the margin of error:
[tex]$$
\text{Margin of Error} = z_{\alpha/2} \cdot \text{Standard Error} = 1.28155 \times 0.0929 \approx 0.119.
$$[/tex]
Thus, the confidence interval is
[tex]$$
1.69 \pm 0.119.
$$[/tex]
In other words, the 80% confidence interval for the population mean is approximately
[tex]$$
(1.57, \, 1.81).
$$[/tex]
This matches the option [tex]$1.69 \pm 0.119$[/tex].
