To construct a 90% confidence interval for the population mean, we follow these steps:
1. Identify the given data:
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 2.86
- Sample standard deviation ([tex]\(s\)[/tex]) = 0.78
- Sample size ([tex]\(n\)[/tex]) = 15
- Confidence level = 90%
2. Determine the degrees of freedom:
- Degrees of freedom ([tex]\(df\)[/tex]) = [tex]\( n - 1 \)[/tex] = [tex]\( 15 - 1 \)[/tex] = 14
3. Find the t critical value (t[tex]\(_{\alpha/2}\)[/tex]):
- For a 90% confidence level and 14 degrees of freedom, the t critical value ([tex]\(t_{\alpha/2}\)[/tex]) is approximately 1.761 (This value is found in t distribution tables or statistical software).
4. Calculate the standard error (SE) of the mean:
[tex]\[
SE = \frac{s}{\sqrt{n}} = \frac{0.78}{\sqrt{15}} \approx 0.201
\][/tex]
5. Determine the margin of error (ME):
[tex]\[
ME = t_{\alpha/2} \times SE = 1.761 \times 0.201 \approx 0.355
\][/tex]
6. Calculate the confidence interval:
- Lower bound:
[tex]\[
\bar{x} - ME = 2.86 - 0.355 \approx 2.505
\][/tex]
- Upper bound:
[tex]\[
\bar{x} + ME = 2.86 + 0.355 \approx 3.215
\][/tex]
Thus, the 90% confidence interval for the population mean is approximately (2.51, 3.21).
So, the correct answer is:
A. (2.51, 3.21)
